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SE 048 835

Blake, Sylvia, Ed.; And OthersFundamentals of Mathematics, Part II. PreparingStudents for the RCT. Experimental Edition.New York City Board of Education, Brooklyn, N.Y. Div.

of Curriculum and Instruction.Curric-00-6090-31; ISBN-88315-634-2Dec 81185p.; For related documents, see SE 048 833-834.

New York City Board cf Education, Division of.

Curriculum and Instruction, 131 Livingston St., Rm

613, Brooklyn, NY 11201 ($8.00).

Guides Classroom Use Guides (For Teachers) (052)

MF01 Plus Postage. PC Not Available from EDRS.

Analytic Geometry; *Computat'Jn; Equations(Mathematics); Graphs; Lesson Plans; *Mathematical

Concepts; *Mathematics Instruction; Measurement;*Problem Solving; Secondary Education; *SecondarySchool Mathematics; Solid Geometry; *Teaching

Guides*New York City Board of Education

ABSTRACTThese materials are intended to provide meaningful

mathematical experiences for pre-algebra students. These experiences

emphasize the development of computational skills, mathematical

concepts,fand problem-solving techniques. This bulletin may be used

as the basis for the second term of a one-year course, or for the

second year of a two-year course. The complete course is divided into

12 basic units, the last six of which are in this bulletin. These

are: (1) solving simple equations; (2) tables, graphs and coordinate

geometry; (3) banking; (4) areas; (5) indirect measure and scaling;

and (6) solid geometry. Each individual lesson includes the aim,

performance objectives, essential vocabulary, specific teaching

suggestions, applications, and a summary. The mathematical skills and

concepts are arranged in a sequential order allowing for the

spiraling of skills among units and semesters. (PK)

***********************************************************************Reproductions supplied by EDRS are the beat that can be made

from the original docur it.********************************************..*************%g**********

U S DEPARTMENT OF EDUCATIONOnce of Educational Research and Improvement

EDUCATIONAL RESOURCES INFORMATIONCENTER (ERIC)

This document has been reproduced asreceived from the person or organization

4) riginating itC' Minor changes have been made to improve

CDreproduction dustily

CN Points of new or opinions stared thisdocument do not necessarily represent officiatOE RI position or policy

"P _RMISSION TO REPRODUCE THISMATERIAL IN MICROFICHE ONLYHAS BEEN GRANTED BY

77ziA4e,--TO THE EDUCATIONAL RESOURCESINFORMATION CENTER (ERIC)."

FUNDAMENTALSPreparing Studere for the RCT

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Experimental Edition, Deczmber, 1 )81

BOARD OF EDUCATION 'F THE CITY OF NIEW 101W

OFFICE OF CURRICULUM DEVELOPMENT. AND SUPPORT DIVISION OF CURRICULUM & INSTRUCTION

2 BEST COPY AVAILABLE

New York City Board of Education

James F. Regan, President

Joseph G. BarkanRichard I. BeattieStephen R. FranseIrene ImpellizzenMarjorie A. Lewis

Robert F. Wagner, Jr.

Nathan Quinones, Chancellor

Charles I. Schonhaut, Deputy Chancellor

Jerald Posman, Deputy Chancellor for Financial Affairs

Louise Latty, Chief Executive for Ir .'ruction

Charlotte Frank, Executive DirectorDivision cr Curriculum and Instruction

Copyrig' :E...)1981, reprinted 1985, 1986by the Hoare Education of the City of New York

Application for permission to reprint any section of this material should he made to the Chancellor, 110 Liv ingston Street,Brooklyn, NY 11201 Reprint of any section of this material shall carry the line. "Reprinted trom fundamental. 01Mathematics Part II by permission of the Board of Education of the City of New York

New York City public schools should order additional copies of this publication from the Bureau of Supplies

ISBN No. 88315.634-2 Curric No (X1-6090-31Other persons and institutions may obtain copies of this'. ublication from the Curriculum Editorial and Produ. lion Unit ofthe Office of Curriculum Development and Support, Room 613, 131 Livingston Street, Brooklyn, Nl 11201 See currentcatalog for price. For information, call 17181 596-4907

3

. .

Foreword

This edition of Fundamentals of Mathematics--Part II is designed to provideappropriate mathematical experiences in the development of computationalskills, mathematical concepts, and problem-solving techniques for pre-algebra students. For some students, this course will contribute to latersuccess in algebra; for others, it represents part of a program in everydaymathematical literacy.

This bulletin may be used either as the basis for the second term of aone-year course, or fcr the second year of a two-year course. The completecourse is divided into 12 basic units, the last six of which are in thisbulletin. Each individual lesson includes:

AimPerformance ObjectivesEssential VocabularySpecific Teaching SuggestionsApplicationsSummary

The mathematical skills and concepts are arranged in sequential orderallowing for the spiraling of skills within units and between semesters.Challenging and practical mathematical experiences are provided to stimulatestudent interest and to help overcome arithmetic deficiencies students mayhave.

The units in this new edition are consistent with those in the New YorkState Bulletin General High Schlol Mathematics published August, 1978, thefirst seven of which make up the curriculum on which the New York StateRegents Competency examination is based. A cross-reference between thetopics of the two sets of units is included. There is a strong emphasison the skills, concepts and problem-solving techniques which are measuredby the Regents Competency Test. The actual Regents Competency items aremarked by asterisks.

Teachers and supervisors using the curriculum bulletin are urged to fill outthe questionnaire (t the back of the book and return it to the Office ofCurriculum Development and Support by the appointed date.

;,/iteil/2

ChARLOTTE FRANKExecutive Director

r

r

Acknowledgments

The curriculum bulletin, Fundamentals of Mathematics: Part IIwas prepared by the Division of Curriculum and Instruction, CharlotteFrank, Executive Director.

Irwin saufman, Director of Mathematics, and Marvin Itzkowitz,Assistant to tLe Director of Mathematics, provided the overallsuperivsion.

The educators responsibly for planning, designing, researchingand preparing this bulletin are: Neal Ehrenberg, Assistant Principal/Mathematics, Hillcrest High School, Rochelle Ehrenberg, Teacher ofMathematics, Bayside High School, and Lorraine Largmann, AssistantPrincipal/Mathematics, Newtown High School.

Final editing and revision were facilitated by: Dr. SylviaBlake, Staff Development Specialist, High School English InstructionalServices; Joseph Gehringer, Teacher of Mathematics, Junior High School65, Manhattan; Marvin Schneider, Assistant Principal/Mathematics,Edward R. Murrow High School, Brooklyn. The manuscript was prepared for

publication by Oi Mei Lee and Shevaun Sloan; cover design by MaureenMaguire.

5iv

Contents

Unit VII. Solving Simple Equations 1

Unit VIII. Tables, Graphs and Coordinate Geometry :53

Unit IX. Banking 79

Unit X. Areas 109

Unit XI. Indirect Measure and Scaling 137

Unit XII. Solid Geometry 161

6I.

UNIT VII. Solving Simple EquationsLESSON 1

AIM: To solve open sentences by trial and error

PERFORMANCE OBJECTIVES: Students will be able to:

- distinguish between a mathematical phrase and a mathematical sentence.

- state whether or not a sentence is an open sentence.

- solve an open sentence using trial and error.

VOCABULARY:

New terms: phrase, open sentence, solution

Review: substitution, variable , verb

CHALLENGE PROBLEM:

Write in symbols the expression "any number increased by four."

DEVELOPMENT:

1. Discuss the challenge problem. Recall that a variable is a

symbol used to represent any number. The expression x + 4 is

called a phrase. Review the definition of phrase meaning a part

of a sentence.

2. Evaluate the phrase for several values of the variable x includingnegative values ( x = 0, 1, 3, 4, -7) Review substitution as

the means of evaluating phrases. (See Unit I, Lesson 2).

3. Have students consider the problem: For what value of x will thephrase x + 4 be equal to 3? Students should be able to determine

from item #2 that -7 is the value of x required.

4a. Translate the problem above into symbols:

x + 4 = - 3

Have students recog..ze the difference between the phrase "x + 4"and the sentence "x + 4 = - 3." Note that the " = " symbolrepresents the verb of the mathematical sentence.

4b. The sentence above is called an open sentence because the valueof x is open for discussion.

5. Do Application A.

6. Pose the problem: "For what value of x, gill x increased oy 4 be

equal to -5?

a. Have students translate this sentence into symbols: x + 4 = - 5.

1

b. Have students determine that none of the values used previouslyresulted in x + 4 being equal to -as follows:

S. This should be demcnstrated

When x = 0 0 + 4 = 4

x = 1 I + 4 = 5

x = - 7 - 7 + 4 = -3

c. Allow students to discover by trial and error that whenx = 9, - 9 + 4 = - S. Therefore, - 9 is the only value ofx which makes the sentence x + 4 = - S a true sentence.

7. Define the solution(s) of an open sentence as the value(s)

of the variable which make the open sentence true.

8. Do Applications B and C.

APPLICATIONS:

A. 1. Indicate by the letter "P" or "S" whether each item below is aPhrase or Sentence:

a. 8 + 7 = 15

b. 6 + 2 = 10 3

c. 18

9

d. the product of 4 anI 6

e. the sum of 12 and 4 is less than the product of S and 6

f. the capital of New York State

g. A meter is the basic unit of length in the metric system

h. five meters

2. Determine which of the following are examples of open sentencesand explain why they are open sentences.

a. He is the president of the United States.

b. Washington, D.C. is the capital of the United States.

c. The capital of New York State is New York City.

d. 10 (25) = 250

e. 10 x = 70

f. 5 + 2 = 10

g. 4 - y = - 6

B. Determine whether the value of x, in the parentheses, is or is not a solution

of the open sentence.

1. 4 x = 24 (x = 6 )

2. 2 x + 7 = 15 (x = 3)

13. -T n = 9 (n = 3)

4. 30 = s SO (s = 80)

15. To- x = 2 (x = 20)

C. Using tr_al and error, find the solution for each open sentence below:

1. a + 4 = 4

2. b + 4 = 0

3. x - 3 = - 2

4. Z - 8 = - 9

5. 7 x = 0

6. 7 x = 283

7. x + ri . 1

SUMMARY:

Review new vocabulary.

CLIFF-HANGER:

If x is a positive integer less than 10, for what values of x isx + 4 78?

3 9

LESSON 2

AIM: To solve equations and inequalities given replacement sets


- categorize a sentence as an equation or an inequality.

- state whether a replacement for a variable makes an open sentencetrue or false.

use a given replacement set to find the solution set of an equationor inequality by trial aad error.

VOCABULARY:

New terms: equation, inequality, replacement set, statement

Review: solution

CHALLENGE PROBLEM:

Explain the difference between the following open sentences:

a) x + 4 = 8

b) x + 4 > 8

DEVELOPMENT:

1. Discuss the challenge problem. Elicit from students that a) has an" = " sign whereas b) has a symbol of inequality.

Define: A sentence using the equality symbol is called anequation. A sentence which uses a symbol ofinequality ( ( or > ) is called an inequality.

2. Consider the problem: If x is a positive integer less than 10,for what values of x is x + 4 > 8?

a. Have students list positive integers less than 10. Thesevalues comprise the replacement set. The replacement setis all possible numbers which may be substituted for thevariable in the open sentence. 1, 2, ..., 9

b. Have students substitute each value from the replacementset for the variable in the open sentence x 4- 4 > 8 anddetermine whether the resulting statement is true or false.(Note: Once a replacement has been made for the variable,the sentence is no longer open. It becomes a statementwhich can be judged true or false).

falseWhen x = 1 1 + 4 8

x = 9 9 + 4 > 8

4

true

I0

c. Have students list all values of x from the replacement setwhich make the statement true. x = 5, 6, 7, 8, 9 Thesenumbers comprise the solution of the inequality from thegiven replacement set.

3. Do applications.

APPLICATIONS:

A. Using integers from 5 to + 5, find the solutions to the followingsentences. (If there are no solutions, write "none.")

1. x + 2 = 5

2. x + 2> 5

3. x + 2 = 10

4. y - 3 = - 6

5. y - 3 = 2

6. y - 3 < - 1

B. Solve (find the solutions for) each of the following open sentencesusing the given replacement set.

1.

2.

3.

4.

6 x = 24

>3-Y- 2

x +1

= 7

3a + 7 < 11

11,

?4,

{64''

(0,

3,

5,

1,

4,

6,

2'

2,

6)

7,

4'

31

8)

7-}4

SUMNARY:

Have students explain the difference between an equation and an in-

equality.

Of the open sentences discussed in this lesson, which type may haveseveral values as a solution?

CLIFF-HANGER:

Solve the equation:

x + 327 = 714

S

i1.

LESSON 3

AIM: To solve equations using addition and subtraction


use the appropriate inverse operation and the addition and subtractionproperties of equality to find the solution to equations cf the formx+ a= b and x- a= b where a and b are integers.

- check the solution of an equation.

- state that addition and subtraction are inverse operations.

VOCABULARY.

New terms: addition property of equality, subtraction propertyof equality, inverse operation, check

CHALLENGE PROBLEM:

Find the solution to the equation x - 25 = 47.

DEVELOPMENT:

1. Elicit from the class that the challenge problem may not besolved easily by trial and error. Therefore, a technique mustbe developed. Introduce the addition property of equality inthe following manner. (Note: A double-pan balance or someother similar visual aid may be helpful in developing thisconcept.)

a. Have students consider the statement 25 - 10 = 15. El'citthat this is a true statement.

If S (or any other number) is added to both sides of thisstatement, what can be said about the resulting statement?[25 - 10 + 5 = 15 + S is also true.]

Similarly, if 10 is added to both sides of the statement,the resulting statement remains true. 25 - 10 + 10 = 15 + 10is also true.

Have students generalize that if the same number is added toboth sides of an equation, the resulting equation willhave the same solutions as the original. Explain to theclass that this is called the addition property of equality.

b. Demonstrate this property with open sentences.

For example: If x = 12, then x + 4 = 12 + 4

or x + = 12 + 5, etc.

Have class determine that all of these equations have 12 astheir solution.

6

2. Demonstrate how the addition property of equality can be used

to solve an equation.

For example: x - 2 = 4

Since the same numl- . 1-.e added to both sides of an equation

without changing 1 _cab or falsity, elicit that 2 may

be added to both sides of tais ,..quation so that the resulting

equation is the soluticr.

x 2 = 4 x - 2 = 4

x - 2 + 2 = 4 + 2 or + 2 = + /

x = 6 x = 6

3. Using the above method, have the class solve the challenge problem.

x 25 = 47

x - 25 + 25 = 47 + 25

x = 72

Since the Iolution was not obvious, how do we know that it is the

correct solution? Indicate that the solution should be checkedby substituting the value for x (72) in the original equation to

determine if the statement is true.

Check: x - 25 = 47

I 72 - 25 = 47?....--,,

47 = 47


S. In a similar manner, dev-!loi-, the subtraction property of equality.

If the same number is subtracted from both sides of an equation,

the resulting equation will ha're the same solutions as the original.

6. Have the class solve the cliff hanger from the previous lesson,

x + 327 = 714, as follows:

x + 327 = 714 x + 327 = 714

x + 327 - 327 = 714 - 327 or - 327 = 327

= 387 x . 587

Have students check the result by suirztitpting the value for

x (387) in the original equation.

7. Do Application B.

8. Pose the question:"How do you determine which property of equality

to use in solving an equation?" Have students see that, since the

solution of an equation means undoing what has been done to the

variable, we must use the inverse operation.

13

9. Do Application C.

APPLICATIONS:

A. Solve each equation using the addition property of equality. Checkyour results by substitution.

1. x- 19 = 43 3. 27 = ; - 412. y - 21 = - 6 4. - 43 = p - 15

B. Solve each equation using the subtraction property of equality.Check by substitution.

1. x + 19 = 43 3. 27 = Z + 412. y + 21 = - 6 4. - 43 = p + 15

C. Solve and check each equation using the addition or subtractionproperty of equality as needed.

1. x - 15 = 7 4. 31 = y 92. x + 15 = 7 5. Z - 37 = 923. 31 = y + 9 6. 92 = Z + 37

SUMMARY:

1. What is the advantage of using the properties of equality overthe trial and error method of solution?

148 I

LESSON 4

AIM: To reinforce solution of equations using addition and subtraction

properties of equality

PERFORMANCE OBJECTIVE: Students will be able to:

- use the appropriate property of equality to find solutions to equations

of the form x ± a = b, where a and b are fractions and decimals.

VOCABULARY:

Review: addition property of equality, subtraction property ofequality, inverse operation, check

CHALLENGE PROBLEM:

Find the solution to the equation x + 5.5 = 6.

DEVELOPMENT:

1. Discuss the challenge problem. Elicit that subtraction is theinverse operation which will undo the indicated addition in this

equation. Stress the importance of rewriting 6 as 6.0 before using

the subtraction property of equality. Review the procedure for

adding or subtracting decimals.

2.

Thus: x + 5.5- 5.5

=

=

6.0-5.5

x

Check the solution

= 0.5

x + 5.5 = 6

Does 0.5 + 5.5 = 6?

6.0 = 6

3. Solve and check:

7 = y + 2-13

a. Elicit that 23 should be subtracted from both sides of the

equation.1

7 = y + 2-T

1 123 - 23

9

b. Review the procedure for subtracting a mixed number from awhole number.

0--,3

=3

23 = - 2iX

43

Therefore: 4.3

= y. Check the result.

4. Do applications giving particular attention to lining up decimalsand changing fractions to a common denominator. Emphasize check-ing each solution AND , if the check yields a false sentence,redoing the problem.

APPLICATIONS:

A. Solve and check:1 3

1. n T T

2. C + .2 = 4.6

3. b - 24 = 9

4. 81

= -1

5. d + 12

=1

6. 2, - 6 = 3.7

7. 12 = p + 1.8

8. t - .5 = 2.4

SUMMARY:

1. How do you determine which property of equality to use to solvean equation?

2. How do you determine which number to add or subtract?

CLIFF-HANGER:

Solve the equation: = 9.

10 6

LESSON 5

AIM: To solve equations of the form ax = b and = b where a and b are

positive

PERFORMANCE OBJECTIVES; Students will he able to:

- state that multiplication and division are inverse operations.

- demonstrate the multiplication and division properties of equality

in numerical sentences.

- use the appropriate inverse operation to solve equations of the form

ax = b and T = b where a and b are positive.

VOCABULARY:

New terms: multiplication property of equality, division property ofequality

Review: addition property of equality, subtraction property ofequality, inverse operation

CH!. LENGE PROBLEM:

Which property of equality should be used to solve each of theequations below:

a. y + 4 = 12 b. y - 4 = 12 c. 4y = 12

DEVELOPMENT:

1. Elicit from the class that (a) and (b) in the challenge problem aresolved using the subtraction and addition properties of equality.In (c) the variable is multiplied by the number 4. To solve

equation (c), it is necessary to "undo" the multiplication byusing the inverse ope/ation.

2. Have students determine that division is the inverse operationfor multiplication.

Example: 4 (6) = 24. Therefore, 6 = 244

3 Based on their experience with the addition and subtraction pro-perties, have students state the division property of equality.

If both sides of an equation are divided by the same (non-zero)number, the resulting equation will have the same sclutionsas the original.

Elicit several numerical examples such as:

4 (6) = 24 4 (6) = 24

4 (6) 24 4 (6) 24

2 2 4 4

Emphasize that it is never possible to divide by zero. Illustrate

12why by asking the class to compute and showing that any suggested

answer must be wrong since it wouldn't pass the multiplication check.12

If --- = *, then * times 0 should = 12, but no number times 0 will = 12.

4. Have students determine a procedure for using the divisionproperty to solve the equation in the challenge problem.

4y = 12

Elicit that since the variable was multiplied by 4, dividing by4 will "undo" this operation yielding the value of y.

Thus, 4y = 12

4y 12

4 4

y = 3

This solution should be checked.

4y = 12

Does 4(3) = 12?

12 = 12


6. In a similar manner, have students state the multiplication propertyof equality: If both sides of an equation are multiplied by thesame number, the resulting equation will have the same solutionsas the original. Apply this property to the solution of equationssuch as:

11; = 8

7 ( P7 = 7(8)

P = 56

Check the solution

8-

56Does 7 = 8?

8=8


12

18

APPLICATIONS:

A. 1. Solve and check each of the foil Ong equations:

a. 7a = 63b. 72 = 9bc. 5c 9

d. 2n = 0.6

e. .05r = 2f. .75 =.5ag. 12x = 2436

2. Write the equation that could be used to solve the followingproblem. Then, solve the equation and check the problem.

If 6 bars of candy cost $2.10, find the cost of one bar of candy.(Use x to represent the cost of one bar of candy.)

B. 1. Solve and check each of the following equations:

a. 11 = 20

b. Tx = 24

d. -Y- = 4 27

e. 8.3 =

c. 12 = f. = 4.1,OS

2. John gave his entire stamp collection to 3 friends. When they

divided the stamps equally, each received 123 stamps. How many

stamps were in the collection? (Use s to represent the number

of stamps.)

SUMMARY:

1. What is the inverse operation for division? For multiplication?

2. What property of equality is used to solve each of the followingand how is that property used?

a) -= 9 (Multiplication property of equality.Multiply both sides of the equation by 7.)

b) 16 = y , 17

c) Sp = 123

d) x + 7 = -3

CLIFFHANGER:

Solve and check:

x-3

= _ 9

13 9

LESSON 6

AIM: To multiply and divide signed numbers


- state and apply the techniques for multiplication and division of signed

numbers to obtain the appropriate products and quotients.

VOCABULARY:

Review: signed numbers, positive, negative, product, quotient, factor

CHALLENGE PROBLEM:

What operatiop is used to solve the equation

x

-3 -9

DEVELOPMENT:,

1. Elicit from the class that, in order to solve the above equation,the multiplication property of equality is used and both sidesof the equation must be multiplied by -3.

The solution will look like this:

31-3)= -3 (-9)

x = -3 (-9)

Alert class that they do not yet know how to find the productof two negative numbers.

The problem will be solved after he class has developed a pro-cedure for multiplying signed numbers.

2. Write the following set of multiplication examples on the board:

(1) 4 (2) 4 (3) 4 (4) 4 (5) 4

x 3 x 2 x 1 x 0 x

a. Have class supply missing factor in 5th example by observing the

pattern of previous factors ( -1 ).

b Have students evaluate first four products (12, 8, 4, 0).

Based on this sequence, what must be the next product?(Since the numbers are decreasing by 4, the next product

is 4. It may be useful to refer to the number line.)

c. Recall that "+4" is another :.ay to write 4.

2014

d. Have students determine the next set of factors:

+ 4 and the product ( - 8 )x - 2

e. Therefore, state the following:

1) The product of 2 positive numbers io a positive number(from the first 3 examples).

2) The product of a positive number and a negative numberis a negative number (from examples 5 and 6).


4. Use the following set of examples to elicit the technique forfinding tne sign of the product of two negative numbers:

(1) + 3 (2) + 2 (3) + 1 (4) 0 (5) 1

x - 4 x - 4 x - 4 x - 4 x - 4(- 12) ( - 8) 71-4) (0) 771761)

Have students state:

The product of two negative numbers is a positive number.


6. Return to challenge problem.

The solution is x = + 27.

Have students attempt to check the solution.

They cannot because they do not as yet know the techniques for divisionof signed numbers. (Use this as a motivation for developing thesetechniques.)

x = _-

Does + -23 = -7

9?

7 Develop techniques for division of signed numbers using the fact that

multiplication and division are inverse operations.

For example: 12 because 12 = 3 x 44

Pose the question + 12 This is equivalent to 12 = x (- 4).-4

Have class determine the missing number (- 3).

8. Have students recognize that the quotient of a positive and anegative number is a negative number.

15

0 ."4.1

9. Use a similar technique to develop a way of finding the sign ofthe quotient of two negative numbers.

Have students state: The quotient of two negative numbersis a positive number.

10. Complete the check of the challenge problem:

-3= 9

Does + 27- 3

9?

- 9 9

11. Do Applications C and D.

APPLICATIONS:

A. Evaluate the following:

1) ( 7 ) 2) - 5

x ( - 7 ) x + 93) 8

x 6

4) (+ .5) (- .2) 5) (- 7.3) (+ 4) 6) (.S) (+ .08)

B. Find each of the following products:

1) (- 12) (- 7)

2) (+ SS) (- 18)

3) (+ 2.5) (- .3)

C. Find the following quotients:

1) 10 ;. (- 2)

2) (- 12) (- 3)

3)+ 125

- 5

D Evaluate each of the following:

1) (-4) (-2) (+3)

2) (+4) (-2) (+3)

3) (-6.4) (-7.2) (20)

4) (- 6) (36.5)

5) (+ 2) (- 3)

6) (

4) 42035

5) - 10010

4) (-20) (-10)

-5

5) -33-11

( 17)

16 22

SUMMARY:

Elicit from the class the following simplified statements:

A. The product or quotient of two numbers with the same signis a positive number.

B. The product or quotient of two numbers with different signsis a negative number.

17

23

LESSON 7

AIM: To solve equations of the form + a x = + b and +L= + ba


- apply the techniques for multiplication and division of signed numbers.

- apply the techniques for equation solving to solve equations of the form

+ ax = + b and + X = + b.aVOCABULARY:

Review: inverse operation

CHALLENGE PROBLEM:

Solve and check. - 2.8x = 5.6

DEVELOPMENT:

1. Use challenge problem to review both the division property ofequality as applied to solving equations and the technique fordivision of signed numbers.

a. Use division property of 2.8 x 5.6

equality and divide both - 2.8 - 2.8

sides of the equation by- 2.8

b. Use the technique for divisionof signed numbers to ob-tain the sign of thequotient. Perform the x = - 2

calculation.

c. Have students check the solution thus reviewing the techniquesfor multiplication of signed numbers.

x = 5.6- 2.8

5.6x

2.8

2. Solve and check:

y = _ 7

3.8

a. Use the multiplication Y = _73.8property of equality and

the technique for multiplicationof signed numbersto obtain (3.8)610 3.8 (-7)

the solution. y = -26.6

b. Have students check thesolution to review the techniquesfor division of signednumbers.

18

24

3. Do applications.

APPLICATIONS:

1. Find the solution

a. 3m = 18

A. 21Z = - 63

g. - S = 3y

. xJ._ 3- = 12

xm. 7- = - 2

1

for each equation

b. 9t = 45

e. - 55 = llk

h. - 9a = - 6

k. = - 72c

m4-1..

and check.

c. 14x = - 42

f. - 5x = - 9

i. Y= -2

x1.-

6= - 1.8

o. - 8k = 8.8n. =- 4 4

SUMMARY:

How do you determine whether to use the multiplication property ofequality or the division property of equality to solve a particularequation?

19 .

25

LESSON 8

t.IM: To reinforce the solution of simple equations


- select which property of equality to use to solve a one-step equation

- solve any linear equation where only one property of equality must beused.

CHALLENGE PROBLEM:

Name the property of equality that was used fo: each equation given inColumn I, to find its solution, given in Column II.

Column I

a. 20m = - 60

b. x + 7 = 12

c. -q- = -93

d. t - 4 = 3.9

Column II

m = - 3

x = 5

q = - 27

t = 7.9

DEVELOPMENT:

1. a. Review the four properties of equality previously developed.

b. Have students state the meaning of each property of equality.

c. Review with students that the inverse operation is used todetermine the choice of property needed to solve the equation

2. Do applications.

APPLICATIONS:

Solve and check each equation.

1. 4x = 32

2. w - 1.6 = .3

Show all steps.

8. 34 = - 8x

9. - 29 = x - 47

3. - 3a = - 21 10. 14 =-

4. x- 18 = - 30 11. = 12-- 4.3

5. y + 5 = 22 12. 6y = - 18

6. s + 18 = 7 13. - 2.6r = 12.74

7.7T= x -

1514.

1 310- = x + 4T

15. x - sT =T

SUMMARY:

Prepare a short quiz on problems of the same type as those shown above.

20

26

LESSON 9

AIM: To solve equations of the form ax + b = c by trial and error


- use substitution to determine whether or not a specific value of avariable is r, solution to an equation of the form ax + b = c or

a + b = c.

VOCABULARY:

Review: substitution, order of operations, statement

CHALLENGE PRCBLEM:

Seven less than 4 times a number is 17. Find the number.

DEVELOPMENT:

1. Discuss the challenge problem. Translate ,.:ifs problem into an tionequation using n to represent "the number."

4n - 7 = 17

Elicit from the class that one method of solving this problem isto substitute different values for the variable until we find thevalue of n which makes the statement true.

For example: If n = 1, 4n - 7 = 17 becomes4 (1) 7 = 17

(It is important tc review the order of operations for evaluatingexpressions. Remind students that multiplications and divisionsare performed before additions and subtractions.)

4 - 7 = 17 is a false statement.Therefore, 1 is not the solution.

Continue substituting values for n.(The students may suggestthese values until the correct solution has been found.) If

n = 6 then 4n 7 = 174 (6) - 7 = 17

24 7 = 17 is a true statement. Therefore, 6 is thenumber that sc.. -es the challenge problem.

2 Have students solve the equation belcw by substituting differentvalues for x.

6 + 3x = 18

(If the student thinks "2" is the solution, an error has been madein the order of operations.) The correct solution is

x = 4 because: 6 + 3x = 18

6 + 3 (4) = 18

6 + 12 = 18 (a true statement)

21 2

3. Do applications.

APPLICATIONS:

A. Determine whether or not the number in parentheses is a solution tothe open sentence:

1.

2.

6n + 17 = 47

55 - 2t= - 32

(n = 5)

(t = - 6)

3. 27 - 2y = 50 (y = 2)

4.x

6+ 2 = 0 (x = 12)

5. + 2 = 24 - 20y (y = )

B. Using the integers from -2 to +2, find solutions for the followingequations. (If there are no solutions, write "none".)

1.

2.

3.

3x + 3

5x - 9

4y + 2

=

=

=

6

1

10

4.

5.

7x

c7 +- 9

6 =

=

5

- 9

C. Using trial and error to choose a value for the variable, find thesolution for each equation.

1. 2c - 4 = 6

2.2b 6 = 12

3. 4a + 2 = - 34

SUMMARY:

1. Review the method of substitution to determine whether or not aspecific value for a variable is a solution to an equation.

2. Review-the order of operations.

CLIFF-HANGER:

Solve the equation:

3x - 5 - 88

2822

LESSON 10

AIM: To solve equations of the form ax + b = c using properties of equality


state that the addition or subtraction properties must be used beforethe multiplication or division properties in the solution of equationsof the form ax + b = c.

- solve and check equations of the form ax + b = c.

VOCABULARY:

Review: inverse operations, order of operations

CHALLENGE PROBLEM:

Solve the equation: 5x + 9 = 79

DEVELOPMENT:

1. Discuss the challenge problem. Students should be led to determinethat trial and error substitution is difficult it solutions toequations are not obvious. A technique for solving equations whosesolutions are not obvious must be developed.

2. Remind students that solving an equation means that we must "undo"everything that was done to the variable. The equation above indicatesthat the variable x was multiplied by 5. 9 was then added to theproduct. What steps are necessary to "undo" these operations?

3. Present the students with a real-life situation in which severaloperations are performed. Determine the order in which the inversemust be performed to undo all the operations. For example: whena man dresses wearing a 3-piece suit, he puts on a shirt, then avest, and then the jacket. To "undo" this, or get undressed, theinverse operations are performed in the reverse order.

4. Return to the challenge problem. Since addition was the lastoperation performed, it must be "undone" first. Then undo themultiplication. Review the inverse operation as the method of"undoing" operations.

5x + 9 = 790 = -9 Undo addition

5x 705 5 Undo multiplication

x = 14

Check: 5 x + 9 = 79

Does 5(14) + 9 = 79?

70 + 9 = 79 v/

23 29

APPLICATIONS:

A. Solve and check numbers 1-10 below:

1. 4n - 6 = 10 6.-5

19 = -24

2. 18 = 2a + 6 7. -3a - 1.2 = 7.2

3. 3y - 5= 8 8. 8.3 = .5x + 6

4. St - 10 = -25 9. :T + 6 = -15

5. -9 x +1 = 10 10. -Y- - 3 45

B Write an equation for each problem using n as "the number."

Solve and check each equation.

1. If ten is added to 3 times a number, the result is 28. Find the

number.

2. Three times an unknown number decreased

ninety-six. Find the number.

3. If seven times a number is increased by

Find the number.

by 12 equals negative

fifty, the result is 15.

4. If ten is subtracted from a number divided by 4, the difference

is 14. Find the number.

SUMMARY:

Have students state the order in which the properties of equality ar-applied to the solution of equations of the form ax + b = c or

b = c.

(Elicit that properties must be used in reverse of the order of operations.)

CLIFF-HANGER:

If two-thirds of a number is 4, find he number

3024

LESSON 11

AIM: To solve equations of the form y = c using properties of equality


- write the reciprocal of any number.

- state that dividing by any number is equivalent to multiplying byits reciprocal.

ax- solve and check equations of the form IT= c using properties ofequality.

VOCABULARY:

New terms: reciprocal

Review: multiplication property of equality, division property ofequality

CHALLEhuE PROBLEM:

Solve the equation:2x = 4

DEVELOPMENT:

1. Discuss the challenge problem. Some students may solve the problemby trial and error, while others will treat it as requiring twosteps (first multiplying by 3 and then dividing by 2). Stillothers may attempt to divide both sides by 2 .

3

2. Consider x = 4 as a 2-step equation:3

2Multiplying) -f x= 4 Check:

2x = 4

3by 3

23 2

= 3 (4)Does -f (6) = 4?

2

2 )(

Dividing ) 2x = 12)rx 4

by 2 ) 2 2 1

x = 6 4= 4

This procedure can be simplified into a one-step process using onlythe multiplication property of equality with the number 3 Thus:

2.2s- x = 4

1 1 2

;i (2"= )

1

1 x = 6

2 5 10, 1

i-2

is called the reciprocal of ,3

Define: Reciprocals are two numbers whose product is 1. Each

number is the reciprocal of the other.

3. Do Applications A and B.

4. Consider32

x = 4 as an equation of the form ax = b. In previous

lessons students learned to "undo" multiplication by using thedivision property of L.quality. Therefore, we may solve theproblem in the following

2x = 4

way:

The right side of the equation may be

2 2 re-written as 4 : 2, Therefore,

32

3

x = 43

The solution is equivalent to evaluating the arithmetic expression

4 1 ,3

W4

rite: T2

T.4 3

Since we know that the solution is 6 and that y X -2-= 6, studentsshould be led to write the statement:

4 . 2 4 3

T T Tx 2*

Elicit the statement: Division by a number is equivalent to multi-

plication by the reciprocal of that number.

5. Do Applications C, D, and E.

APPLICATIONS:

A. Write the reciprocal for each of the following:

1)

2)

3)

4)

S)

3

1-s

5

11

2- g-

6)

7)

8)

9)

10)

1-

-8

-

5T

.3

3226

B. Solve each of the following equations using the multiplicationproperty of equality and reciprocals.

1)3y = 12

4

12) m = - 23

3) 17 = Sx

1 14) 1Ty =

5) - 40 = - t5

6) - rifr = 4

71 - 8q = 2

1 58) - 2n = T3

1 59. - 2-

3= y

10) . 3x = 15

C. Perform the indicated operations and express the results in lowest terms.

.

71

4) 334 11T

5) 30

2 . 36) -s 7

16

D. Solve each of the following equations using the division property ofequality.

1) -5x = 12

12) Tn = - 40

3) sx = - 20

4) 4y = 33

5) - S4 = 3x

E. Solve each of the following:

1 If a board 104 long is cut into 4 equal pieces, how long4

will each piece be?

2 A SO pound bag of peanuts is to be placed into small bags holding

16of a pound each. How many smaller bags can be filled?

SUMMARY:

1 Review the meaning of the terms "reciprocal" and "finding thereciprocal of a number."

2 Review the method used for division of fractions.

3 Review the methods for solving equations of the form 22i=b

c.

CLIFF-HANGER:

If -1 of a number is increased by 7, the result is 16. Find the number.4

27

"3

LESSON 12

AIM: To solve equations of tne form -P-1 + c = db


- state the sequence of steps required for the solution of equations ofax

the form IT.2 c = d using inverse operations.

- solve the equations and check the solutions.

CHALLENGE PROBLEM:3

Solve the equation: Tiy + 7 = 16

DEVELOPMENT:

1. Review the procedure for solving equations where several operationsare performed. Recall the development from Lesson 10 in whichoperations must be "undone" in exactly the opposite order fromthat in which they were originillly performed.

3Ty + 7 = 16

Subtract 7 from both sides. - 7 = - 7

4y = 9

Choose one of the methods developed in the previous lesson to3

solve Ty = 9.

Method 1 or Method 2

3

1 9 4 9

4 3 ig3x -4-y

4x

y = 12

Check the result.

2. Do applications

28

3

TIY = 9

3

4 4

. 3Y = 9 7 T

3A"4

y = x

y = 12

APPLICATIONS:

A. Solve and check the following equations:

1) -2-m 6 = 15

2) 9 = -18.; 11

23) Ty - 3 = 9

24) s + 7 = 29

3

35)

4b 14 = 8

B. Solve each of the following:

6) - 25 = 3s - 11

5d7) -T * 12 =82

8) 4 = -5-6t

21) A number is multiplied by T. When this product is decreased by 13,

the result is 7. Find the number.

7 3 12) IT of a number decreased by is 4T Find the number.

53) If 38 ,s added to .§rof a number, the result is 128. Find the

number.

SUMMARY:

axReview the procedure used to solve equations of the form -6 c = d.

29

LESSON 13

AIM: To review the solution of linear equations using one or more propertiesof equality


- use formulas to solve problems by substituting given values andsolving the resulting equations.

- determine if the solution satisfies the conditions of the problem.

VOCABULARY:

Review: formulas, and all terms related to solution of equations.

CHALLENGE PROBLEM:

F =9C+

32 is a formula relating Fahrenheit and Celsius temperatures.FindFind the Celsius temperature when the Fahrenheit temperature is 85 0 .

DEVELOPMENT:

1. Have students recall the procedure for substituting values ina formula.

Elicit from the class that once the substitution is performed,an equation with one variable is obtained.

F =2C+ 32

5

985 = 5 C + 32

Using substitution, review the procedure for solving this equation.

Subtract 32 fromboth :ides.

85 -9 C + 325

-32= -32

9 (;

C

Multiply both

sides by9

53 =9

; (53)

265

C

=

=

=

9

294

40The temperature is 2g-- c.

9

Check the result.

C

2. Provide students with a variety of equations and verbal problemsreviewing all techniques presented in this unit. Some interestingapplications follow:

30 36

APPLICATIONS:

Use the given formula to solve each problem. Check your results.

31 A car was driven a distance (d) of 51 kilometers in -a- of an hour

(t). Find its rate of speed (r). (d = r t)

2. The retail price of an item is the sum of the wholesale price and

the mark-up (r = w + m). A jacket has a retail price of $37. A

store's mark-up is $17.25. What is the wholesale price?

3. The sale price of an item(s) is found by subtracting the discount

(d) fron, the regular price (r). What was the regular price of an

item that was on sale for $12.74 if the discount was $2.25?

4. The formula relating a person's blood pressure and age is

p =a

+ 110. Sandra's blood pressure is 134. How old should

she be?

5. A person's height in inches and weight in pounds are approximately11

related by the formula w = --h - 220. According to this formula,

what is the height of a person weighing 220 pounds?

UNIT VIII.Tables, Graphs, and Coordinate Geometry

LESSON I

AIM: To read and interpret data found in tables


- provide examples in which presentation of data in tables is useful.

- read tables of data and answer specific questions relating to

these data.

- describe ways of presenting data: lists, tables, charts, maps, graphs.

VOCABULARY:

New terms: tables, charts, graph, horizontal, vertical

Review: data

CHALLENGE PROBLEM:

List at least five examples of information commonly presented in tables,

charts or graphs.

DEVELOPMENT:

1. Students will probably suggest examples such as:

sales tax, train or bus schedulec, hospital charts, T.V.schedules, postal rates, weather charts, ideal weight chart,sports statistics, stock market summary, travel maps, etc.

2. Elicit from students that tables and graphs are used to present

data in a convenient, easy-to-read way. The form used for the pre-

sentation depends upon the information given and the intended use.

3. Consider information that is often presented in table or chart

form. One example is a mileage chart.

Distances Between Cities in New York State

=M.n

MC4.1

t4.

z=

>- 4-)

::: u0

-.?:

M .-1,-1

m m1 r-:-.x

.=.4=b+5

R

0Ms,>-,

JD

$'4

4-3

m

Albany 291 154 300 71 142 173

Buffalo 291 397 22 320 146 200

New York City 154 397 - - 419 72 251 323

Niagara Falls 300 22 419 372 165 220

Poughkeepsie 71 320 72 372 4S 298

Syracuse 142 146 251 165 45 --- 71

Watertown 173 200 523 220 298 71

Distances Given in Miles

33

a) What is the distance between Niagara Falls and Poughkeepsie?Discuss the two ways of finding this answer. (Note horizontaland vertical labels.) Allow students to discover the symmetryof this table and compare that with symmetry in addition and/ormultiplication tables.

b) Consider other questions relating to this table such as:

i. Which two cities are the closest together? Furthestapart?

ii. If the speed limit in New York State is SS miles perhour, what is the minimum travel time between Syracuseand Niagara Falls? etc.

iii. On a business trip a company provides a travel allowanceof 21(t a mile. What is the allowance for a trip fromNew York City to Albany? A round trip between Watertownand Poughkeepsie?

4. Do applications using the samples presented or more current tablesand charts found in newspapers, magazines, etc.

APPLICATIONS:

A. Use the telephone rate table shown to ansver the questions below.(Note the difference between the presentation of the table her and theone used in the development. Why is only "half" of a table necessary?)

316 300

3 0013 00 316

3 1613 16 3 1612 99 3002 5012 66 2 50(3 00 3.00

3 16 3 3 3 1 3 33 3 3 3 ! 3 t 6 3 1 6

3 1613 3313 33 3 33 i 3 16 3 16

301 3 II

3 0013 16

300 3 001 3 003 00 3.001 3 16

3 331 1 33 3163 1613 331 3 16

3 16 2 99 3 00

316 300 300316 299 3003 00 3 00 12 99

316 299 3C0316 3 33 3163 '613 33 3 16

263 3 002 SO 3 00

2 66 3 003.111 3 00

2 63 3 003 33 3 33

3 33 13 33

1. What is the cost of a 20-minute call between Dallas andPhiladelphia?

2. What is the total cost of three 20-minute calls between New YorkCity and Boston?

34 :39

B. Use the sales tax table shown to compute the total cost of the purchasesof taxable items in a local candy store.

$0.01 to $0.10 00 3.07 to 3.18 250.11 to .17 10 3.19 to 3.31 26C.18 to .29 2C '3.32 to 3.43 27C.30 to .42 3C 3.44 to 3.56 28C.43 to .54 3.57 to 3.68 = 290.55 to .67 - 5C 3.69 to 3.81 30C.68 to .79 - 6C 3.82 to 3.93 - 310

to .92 - 7C 3.94 to 4.06 320.93 to 1.06 PC 4.07 to 4.18 33C

1.07 to 1.18 9C 4.19 to 4.31 - 3401.19 to 1.31 100 4.32 to 4.43 35C1.32 to 1.43 - 110 4.44 to 4.56 36C1.44 to 1.56 120 4.57 to 4.68 37C1.57 to 1.68 130 4.69 to 4.81 - 38C1.69 to. 1.81 - 14C 4.82 to 4.93 - 39C1.82 to 1.93 150 4.94 to 5.06 - 4001.94 to 2.06 - 160 5.07 to 5.18 4102.07 to 2.18 17C 5.19 to 5.31 42C2.19 to 2.31 180 5.32 to 5.43 - 43C2.32 to 2.43 19C 5.44 to 5.56 4102.44 to 2.56 200 5.57 to 5.68 '4502.57 to 2.68 21C 5.69 to 5.81 46C2.69 to 2.81 22C 5.82 to 5.93 - 47C2.82 to 2.93 23C 5.94 to 6.06 4802.94 to 3.06 24C

1. A game for $5.99

2. Three birthday cards at 60* each

3. The following schools supplies:

1 pack loose-leaf paper - 99¢

2 ball-point pens @ 39* each

1 spiral notebook $1.19

6 pencils at 10* each

35

Long distance rates within New YorkDIRECT DISTANCE DIALED

(Paid By Calling Party)

FROM

MANHATTANTO:

DAY

8 AM to 5 PM Mon -Fri.

EVENING

5 °M to 11 PM Mon.Fri.

Weekends 8 AM to 11 PM

NIGHT

11 PM to 8 AM All Days

Init,2

Mins.

Ea.

Add.Min.

2

(Ans.

Ea.

Add. 1

M;n.

Ea.

Add.Min.

Albany S .90 S.40 S.58 S.26 20 S.16Binghamton .90 .40 .58 .2o .20 .16Huntington .52 .22 .33 .14 .12 .08Monticello .80 .35 .52 .22 .18 .14Montauk Point .90 .40 .58 .26 .20 .16Mt. Kisco .52 .22 .33 .14 .12 .08Niagara Falls 1.02 .45 .66 .29 .22 .18Poughkeepsie .73 .31 .47 .20 .16 .12Riverhead .80 .35 .52 .22 .18 .14qnchester 1.02 .45 .66 .29 .22 18Syracuse 1.02 .45 66 .29 .22 .18Utica 1.02 .45 66 .29 .22 .18

C. Use the tr e of Long distance Telephone Rates within New York Tocompute the cost of telephone calls from Manhattan.

1. A 2-minute telephone .all to Huntington at 9:00 A.M. on Monday?at 9:00 P.M.? at 7:45 A.M.?

2. How much more does iv cost to place a 5- minute call between NewY'rk City and Mt. Kisco on a Friday afternoon (before 5:00 P.M.)tnan on a Saturday afternoon?

41_36

D. Use the Guide to Street Numbers in Manhattan to answer the fc owing:

GUIDE 10 STREET NUMBERS IN MANHATTANTo End nearest street. drop last dgit of house number

dhode by 2 and lunher calculate as shown belowAve A B C. 0 Add 3 Audubon Ave Add 1651st 6 2nd Ave Add 3 Broadway Up lo 7543rd Ave Add SO ,S below 8th St4th Ave 754 to 858 Subt 29

(Park Ave S) Add 8 858 to 958 Sub1 255th tve Above 958 Subt 31

Up to 200 Add 13 ColumbuS Ave Add 60Up to 400 Add 16 Convent Ave Add 127Up to 600 'Add 18 Ft WashingtonUp to 775 Add 20 Ave Add 158776 to 1286 Lenox Ave Add tio

canc.; loll ltaufe Lexngton Ave Add 22and sub: 18 Mathson Ave Add 26

Up lu 1500 Add 45 Manhattan Ave Add 100Above 2000 Add 24 Park Ave Add 35

Ave of the Si NieholaSArneneaS Subt 12 Ave Add 110

71h Ave west End Ave Add 60Up to 1800 Add 12 Central Park WestAbove 1800 Add 20 Otnde house number by

8th Ave Add 10 10 and add 609th Ave . Add 13 Pedersode Drive10th Ave Add 14 ()wide house number by11th Ave

A10 and add 72

Amsterdam Ave Add '560 up to 165th St

SUMMARY:

1. What is the closest street to147 Fifth Avenue? 666 Fifth Avenue?

Between which two sLreets is878 Broadway?

Have students describe situations in which tables are used to presentinformation.

37x2

LESSON 2

AIM: To read and interpret pictographs (picture graphs)


- read the scale on a pictograph.

- translate the symbols on a pictograph into numerical quantities.

- extrapolate new information or form opinions using the data on apictograph.

VOCABULARY:

New terms: pictograph, symbol, legend, scale

CHALLENGE PROBLEM:

If the symbol represents 10 hits for a baseball player, draw the

representation for 55 hit'.

DEVELOPMENT:

1. Elicit the response to the challenge and discuss how the answer isobtained.

2 Have students rec 1 that data may be presented in many forms.One form that is usually easy to read and visually attractive isa pictograph. The pictures in a pictograph often convey the sub-ject matter of the graph.

(Note: Appropriate gra- hs should be prepared for overheadprojector or rexograph , etc.)

3 Present a pictograph similar to the one shown below and discussthe components of a pictograph. (Note: This graph is incomplete;items below explain how the graph should be completed.)

MICHAEL

JOHN

CARLOS

TROY

38 4

4. a) What information can be obtained from this graph? Elicit thateach row of symbols is !receded by a label. This label canbe com,ared to the vertical labels on some tables.

b) What does this graph show? Elicit from students that a titleis necessary in order to answer this question. Provide thetitle: Leading Hitters at Eastern High School, and repeat thequestion.

5. Which player on the team had the most hits? How many hits didthis player have? Elicit that further information is necessary.Every pictograph must have a scale or legend which indicates thevalue of each symbol. Provide the information = 10 hits.Repeat the question and provide additional questions relating tothis graph such as:,

a) How many more hits did Troy have than Carlos?

b) How many more h'ts does the leader have compared to the 2ndplace hitter? 3rd place hitter?

c) If this chart represents hits for the first half of the year,how many hits would you expect Michael to have at the end ofthe year if he continued hitting at his present rate?

6. Do applications.

APPLICATIONS:

A. Ice-cream cones sold at Steve'b 7 Flavors on July 4.

iTh (7

Rock y Road VPink Bubbi

Peanut Bu. en

Pina Colada

Strawberry Ripple (.9 (9

Chocolate Mousse

Vanilla Bean

(Th

= 20 cones

39

1. Which flavor was the most popular? least popular?

2. How many cones of Pink Bubblegum ice-cream were sold? Pina Colada?Rocky Road?

3. How many cones were sold on July 4?

4. If this pattern continues, how can Steve use the information toplan his production?

B. Points Scored in the Play-Off SeriesBy the Starting-Five Basketball Players

FRANKLIN

DOUGLAS

MONROE

DEREK

STRETCH

Legend: = 16 points

How many points did each player score?

C. If each symbol represents 10,000 records sold, how many records arerepresented by each row?

a)

b)

SUMMARY:

1. What information is needed in all pictographs?

2. If one symbol represents 4,000 cars sold, how many symbols are neededto show 300,000 cars?

3. Would you consider this method of :;hawing data very accurate?

40

45

LESSON 3

AIM: To read and interpret bar graphs


- identify bar graphs as a type of graph that can illustrate data moreaccurately than pictographs.

- translate the scale units used on a bar graph into numerical quantities.

- read and interpret bar graphs.

read and interpret bar graphs with double sets of data.

VOCABULARY:

New terms: horizontal and vertical scale (axes), bar graph

Review: legend, pictograph, symbol, scale

CHALLENGE PROBLEM:

If each ()represents 15 points, how many points are represented by

DEVELOPMENT:

1. Elicit from the class that the solution to the challenge problemis difficult because it is hard to estimate the :racLional partre2resented by the figure. Similar problems arose in the previous

le ;son. Have the class determine that a more accurate means ofgraphically presenting data is desirable.

2. One such graph, closely related to pictographs, is called a bar

graph. Present the following:

VANILLA

NIDE)E1 OF VALE-GALLLOSS Of ICE CREAM WLD IN JCWSDELI DURING Tor rom or JULY

crocourz 7-7

nutatirm F---11

pampa, [7--]

CI I

Noce: Ono So equals Can Lit-gilloos of ice creme

B.

41

VAMILLI

CDO:OIATE

STRAWBERRY

?MACY:0

NUMBER Or MALF-GALLONS OF ICE CLEARS SOLD DI .30I01'S

ELI DVR vG ItE WWII OF JVLT

LIKI

NV I 1 \MUM .6

MI LI &WM IMN MKS

.0

4 6

20 30 40 SO

!AL -GALLCRS

Compare the two graphs and elicit the following:

a) Both graphs have titles and an indication on the verticalaxis telling what each row of symbols or bars represents.

b) The scale for a pictograph is ,ymbol, but the scale for abar graph is a length on a line

c) To find the number of items in a pictograph you must count thenumber (or parts) of symbols and multiply by the value of eachsymbol. To find the number of items represented by a bar ina bar graph, estimate where the end of the bar falls in rela-tion to the scale.

3. Elicit that the differences make the bar graph easier to read.(However, a pictograph can be visually more attractive.)


5. Elicit that in the bar graph presented, the bars were drawnhorizontally and a number scale was on the horizontal axis. Thevertical axis contained the labels. It is also possible to drawthe bars vertically. In this case, where must the number scale beshown? What information must be shown on the other axis?

6. Present the following vertical bar graph. Discuss its parts andanswer the following questions:

ATTENDANCE AT A CONCERTONE WEEK IN AUCUST

w7

La 60..1;

o c4

(;)

0 .%CD=M"2D

MON TUES WED THUR FRI SAT

a) On which day did the most people attend? Fewest?

b) Approximately how many people at !Tided on Wednesday? Thursday?

c) What is the difference between the largest crowd and thesmallest crowd?

d) What was the approximate total attendance for the week? Theaverage daily attendance?

7. Do Application U.

8. Present the table and multiple bar graph that follows. Comparehow the same information can be presented in two different ways.From which presentation is it easier to extrapolate information?

42 47

The Weather in Five Cities

B Cit C City D City E

Clear 101 days 6 days I 72 days 84 days 223 days

Cloudy 166 days 204 days l 78 days 167 days 54 days

Rain or Snow 109 days 154 days 132 days 87 days 28 days

220

200

180

, 16003

:'! 140

ti 120av., 100co

C:t 80

6040

20

The Weather in Five Cities

0 City A City B City C

Key: Clear ri Cloudy

City D City E

Rain or Snow

APPLICATIONS:

A. Answer the following questions based on the given bar graph:

Alice's Commission for 5 Months$900

$800

$700

$600

$300

$200

$100

0

43

48

1. What was Alice's total commission for January? For February?

2. For which month was Alice's commission highest? Lowest?

3. For which two months was it the same?

B. Major Causes of Fire in New York City

Heating or Cook-ing Equipment

ElectricalEquipment

Open Flames orSparks

Arson

Explosions

0 50 100 150 200 250 300 350 400 450

Number of Fires

1. What are the two major causes of fires?

2. About how many fires did electrical equipment and heating orcooking equipment cause?

3. About how many times more fires are caused by explosions than arecaused by heating or cooking equipment?

4. Approximately how many fires were caused by arson?

C Use the double bar graph to answer the following questions:

Student Scores on Math Tests

100

90

80

6- 70

0O 60

O 50

, 40043E 30

20

10

0Inn Bob Carlos Kim Lance Lee Marco

Student

sm

k a .q r 1 k

% \ b11

,. SI§

!I t ISM 1 I IS1 1la

11111 INIS ER11111 SS IS 11

Al5S111115 ll It II51 1 N 1 ANN iS7

Pre- Post-

test N test

44

49

1. Who scored the most points on the pretest? The fewest points?

2. Who scored the most points on the post-test? The fewest points?

3. Who scored 30 points higher on the post-test than on the pretest?

15 points higher?

4. Who scored the same on the pretest and post-test?

5. Which students had the same scores on their post-tests?

6. Which student showed the most improvement from pretest to post-test?

SUMMARY:

List the major components of a bar graph (title, horizontal and vertical

scales or labels).

45

LESSON 4

AIM: To read and interpret line graphs


identify line graphs as types of graphs that illustrate trends moreeffectively than bar graphs.

- read and interpret 1.11e graphs.

- state that line graphs tend to show a continuous series, process, ortendency (optional).

VOCABULARY:

New terms: line graph, trend, (interval)

Review: horizontal and vertical scales (axes)

CHALLENGE PROBLEM:

Howard kept track of the weekly closing of his stock for six weeks andfound the following data:

Week Closing Price (in dollars)

1

2

3

4

5

6

30

3

33 T1

371-

36

42

1

39 T

During which weeks did his stock rise in price?

DEVELOPMENT:

1. Discuss the challenge problem. Elicit that stock rose from week 1through week 3 and week 4 to week 5.

2. In what way might this data be presented so that it may be inter-preted more easily? (a graph) Data of this type is usually pre-sented using a line graph. (Teacher may wish to present linegraphs from newspapers.) Present the following line graph:

61

Howard's Stock Performance

.1 2 3 '1 5

Week

b

3. Discuss the major omponents of a line graph,

a) Title

b) Horizontal and vertical axes each having a scale,

c) Each point (dot) represents 2 pieces of info.lation the week

number and its corresponding closing price,

d) A line segment is drawn between each pair of consecutive dots.Each line segment shows the trend for the indicated period.

4 Do applications.

S Compare bar graphs and line graphs, by presenting a bar graphdrawn from the data in the challenge problem.

47

1.

44

Howard's Stock Performance

42

34

41 - _ 30

0

V/

=3f v 24

=

r/a135 0A0

.3

.7

12

32

1 2 3 4 5 4

Week

Howard's Stock Performance411111,

1

eieek

Elicit the following comparisons:

a) Both graphs have the same title and horizontal axis.

b) On the bar graph, the vertical scale begins at zero whereason the line graph, we chose to begin at 29. Why? Could 31

be used as the starting point?

c) Show that by placing a dot in the center of each bar andconnecting the dots consecutively, a line graph is obtained.Why do the two line graphs appear different? (Different scales.)

On which graph are the trends more easily sc....? (line graph)

Which graph shows a truer picture of the stocks' performancein relation to its overall cost?

(Teacher may wish to compare this stock's performance to theperformance of another stock whose prices range from 230 to

2391/4.)

6. Present and discuss multiple line graph shown in Figure 3:

4853

140

330

Weights of Two Gyfu.sr r-

PlAik IA

-- -- 1...,10 '1

x,90

se"

8010

ir

11 12 13 14Age in Years

15 16

Study the graph in Fig. 3 and answer the following questions:

a) What type of graph is pictured?

b) What is the subject of the graph?

c) What is shown on the horizontal axis? On the vertical axis?

d) Whose weight is represented by the dotted line graph?

e) At age 15 what was Maria's weight? What was Linda's weight?

f) At which ages did Maria weigh more than Linda?

g) Between which two ages did Maria's weight increase most?

h) At which age was the difference be-ween Linda's weight andMaria's weight greatest?

7. (Optional) Using graph in Application A, pose the followingquestion: Approximately what was the temperature at 2:30 P.M.?Discuss the fact that the lines in a line graph may representuniform change in an interval.

APPLICATIONS: Outdoor Temperature (Celsius) in Newtown, Oct. 30

22°

Figure 4 20°

IS

16°

14°

12

.:4 ...:a: o: o:

54

A. Study the line graph in Fig. 4 and answer the following questions:

1. What is the title of the graph?

2. What does the horizontal axis show?

3. What does the vertical axis Show?

4. At what temper -cure does the scale startf

5. On a line graph, must the scale start at zero?

6. What was the temperature at noon? At 3:00 P.M.?

7. At what hours was the temperature highest? Lowest?

8. Between which hours was the temperature rising? Falling?Unchang '

9. He much did the temperature rise between 11:00 A.M. and noon?Between noon and 1:00 P.M.?

P. Between which two successive hours did the temperature fall /1strapidly?

Performaace of JCN ,fflpany Stock

$ 90

Figure 5

S 70

$ 60

$ 50li v;

..rii d

w 0 IL 0 T

p- t--c

B. Study the line graph in Figure 5 and answer he following questions:

1. What is the decrease in price of the stock between Wednesday andthe fo'lowing Tuesday?

2. What was the stock's price on Thursday?

50

3. Between which two days did the stock fall most in price?

4. What was t'.e stock's price at tLe close of business on the last day?

S. Between which two days did the stock fall least in price?

b. ?that is the trend in the stock price?

Figure 6

Earnings of Harold and Paul

140

= 120Ao X00

80z

60

40Jan. Feb. Mar. Apr. May Jun.

Month of the Year

r0 iA

4 1---lic,Pa.RI

WANVAIIIIM

C. Study the line graph in Figure 6 and answer the following questions:

1. What is the title of the graph?

2. How is the horizontal axis labeled?

3. What scale is used on the vertical axis?

4. Whose earnings are represented by the solid line graph?

S. In which month were the earnings greatest for Harold? For Paul?

6. In which month were the earnings least for Harold? For Paul?

7. What was the greatest amount earned by either of the boys?

8. What was the least amount earned by either of the boys?

9. During how many months were Paul's earnings greater than Harold's?

10. In which month did Harold earn the same amount as Paul?

11. In which month was the Oifference between Paul's earnings andHarold's earnings greatest?

SUMMARY:

Review the purpose and interpretation of line graphs. Emphasize that ttlechoice of scales can distort or cxaggcrate the facts being shown. Onemajor purpose of a line graph is to indicate trends.

LESSON 5

AIM: To read and interpret circle graphs


- identify circle graphs as good graphs to use when illustrating thesubdivisions of a whole quantity.

- read and interpret circle graphs where the subdivisions of a wholeare given as numerical quantities.

- compare the relative sizes of sectors of a circle by inspection

VOCABULARY:

New terms: circle graph

Review: sector

CHALLANGE PROBLEM:

John received $4.00 for his weekly allowance. He spent $1.00 for school

snacks and $3.00 for movies. Which diagram correctly represents thisinformation?

DEVELOPMENT:

1. Using the challenge problem, elicit the fcllowing:

The circle represents John's total allowance. The divisionsof the circle represents the approximate portions of theallowance allotted to each activity. What part of his allow-once was spent on snacks? (%) Which diagram show snacks as

of the whole circle?

... Graphs using the diagrams presented above are called circle graphs.

a) In a circle graph the whole circle represents the entire quantitydiscussed (allowance). "The teacher may present other circlegraphs obtained from newspapers, income tax instructionbooklets, magazines, etc.] The title of the graph indicateswhat the whole circle represents. What might be an appropriatetitle for a graph of the information in the challenge problem?

b) Each sector of the circle represents r subdivision of the wholequantity and is labeled appropriately. The size of each sectorrepresents the part of the whole taken up by a subdivision. In the

52 57

the challenge problem, what are the subdivisions? How muchof his allowance is spent on each activity? Enter this in-formation on the diagram. Elicit that the sum of thesequantities is equal to '--.00 (the whole allowance). Thegraph is now complete because:

(1) it has a title, and

(2) each sector has a label and an indicated size.

3 Dr applications.

APPLICATIONS:

A. Sources of income for a local school district:

PikivATe GRANTS

List from which sources the money is received in the order of largestto smallest.

B. Health Education Recreational Activity Choices:

Fig. 3

853

C.

1. How many pupils were in the health education class?

2. What fractional part of the number of pupils in the class selected

basketball? Track? Baseball? Swimming?

3. Which sport was selected by the greatest number of pupils? The

smallest _amber of pupils?

4. How many more students selected baseball than track? Selected

basketball than swimming?

5. What is the ratio of the number of pupils who selected baseball

to the number of pupils who selected track? Who selected basket-

ball to the number of pupils who selected swimming?

How Wai-KeunR Spends EachDollar He Earns

1. How much of each dollar is spent on rent? Food? Car? Clothing?

Other?

2. On which item is the most money spent? Least money spent?

3. How much mnre is spent on rent than on clething?

4. What part of each dollar is spent on rent?

SUMMARY:

What kind of information is represented by circle graphs?

How does the size of the sector relate to the size of the whole?

CLIFF-HANGER:

In Application C, if Wei-Keung earns $600 a month, how much does he

spend monthly on car expenses?

54

59

LESSON 6

AIM: To define percent and find percent of a number


- express percent as the ratio of a number to 100.

- find the percent of a number by expressing the percent as a fractionand multiplying.

- state that the sum of all the parts of a quantity is 100%.

VOCABULARY:

New terms: percent

Review: circle graph, ratio

CHALLENGE PROBLEM:

How Carl Spends Each Dollar He Earns

If Carl earns $600 a month, how much does he spend on rent?

DEVELOPMENT:

1. Discuss the challenge problem. Recall that each sector in a pic-

tograph relates the item shown to the whole quantity. In this case4040* of each $1.00 is spent on rent. This can be expressed as100

of the money spent. Therefore this problem can be solved by finding40100

of $600.

40 600T-515 x

1

120

2 -6ee- 40 600-6

x T o. x1

= 240 = 240

55

402. Consider the fraction

100'Fractions with denominators of 100

are often written in 14-P form 40%. Have students observe the

similarity between the symbol % and its meaning ,,1051, (division

by 100).

Define a percent as the ratio of a number to 100.

40 65For example: = 40% 65% =

100 100


4. Return to the graph in the challenge . Express each sector label23

as a fractional part of a dollar. That is Car, Food,100' 100'

etc. What is another way in which each of these fractions could

have been presented? (As percents, i.e.. 116, 23%, etc.)

5. Explain that many circle graphs present information usii.g

percents. Provide students with examples, such as:

Where Money FromRecord Sales Goes

COMPANYPfkoP IT

25%

MAN d FACTURE

15%Disrateur1c4

AN°SALES

a) Where does the smallest part ofthe money go? Largest part?

b) What percent is paid in roycltiesto the song writer, singer and/ormusicians?

c) What fractional part of themoney from sales is used forma,Lufacturing?

6. Suppose a record album costs $8. How much of this cost is returnedto the company as profit? Have students express this problem as:

25% of $8

Since 25% can be written as a fraction, this problem can be Px-25

pressed in the same form as the challenge problem, that is100

of $8. Solve the problem.

7. Generalize a procedure for finding the percent of a number:

a) Express the percent as a fraction.

b) Multiply the fraction by the given amount and express theanswer in simplest form.

56 61.


APPLICATIONS:

A. 1. Express each of the following as a fraction in lowest terms.

a) 25%

b) 30%

c) 20%d) 7%

e) 100%f) 15%

g) 16%h) 72%

2. Express each of the following as a percent.3, 4a) e) 5-100

b) 1 to 100 f)3

8

c) c) 100 3 1g) (note: Reduce to first)

d)3

10 h) 1

b. Find the indicated percent of the given numbers.

1. 68% of 100

2. 10% of 750

3. 24% of 75

4. 7% of 600

5. 50% of 362

6. 75% of 36

7. 100% of 721

8. 1% of 2000

C. Accidents to Children in Viewridge

25%

Other places

20%School

grounds

1. What percent of the total number of accidents occur going to andfrom school? On school grounds?

2. Where do the greatest number of accidents to school children happen?

3. How many times more accidents happen in the home than happen inschool buildings?

57

4. What fractional part of the accidents to school children happenat home? In School buildings? On school grounds? Going toand from school?

5. If there were 480 accidents to school children in one year, howmany accidents happened at home? In school buildings? On schoolgrounds? Going to and from school?

SUMMARY:

1. In a circle graph, what percent is represented by the entire circle?

2. Review the procedure for finding the percent of a number.

CLIFF-HANGER:

Using the "Record Sales" circle graph, find the cost of manufacturinga record album that retails for $8.90.

586 3

LESSON 7

AIM: To express percents as decimals and to find a percent of a number


- express whole number percents as decimals.

- find a percent of a number when the percent is expressed as a decimal.

VOCABULARY:

Review: decimal, percent

CHALLENGE PROBLEM:

Find 15% of $8.90

DEVELOPMENT:

I. Discuss challenge problem and compare it to the cliff-hanger.

15 $8.90100

x1

Why is this problem more difficult than the ones done previously?(This problem is not easily solved using fractions.)

2. Elicit that another way of expressing 15% is .15 (read 15 hundredths).


4. Return to challenge problem. Solve as follows: 15% of $8.90.

.15 x $8.90 or $8.90x .154450890

$1.3350 --> $1.34.

Review multiplication of decimals, placing decimal point, and

rounding off to nearest cent.


APPLICATIONS:

A. 1. Express each of the following as a decimal:

a)

b)

c)

d)

42%

13%

5%

1%

e)

f)

g)

h)

100%7

fl5

2

5

1

i

G 159

2. Write each of the following as a percent:

a) .47 e) .06

b) .86 f) .6

c) .30 g) 1

d) .3 h) 1.35

B. Compute the following:

a) 390 of 671 d) 98% of 79

b) 2% of 36 e) 65% of $12.50

c) 6% of 150 f) 82% of 9.9

C. Answer the following questions based on the given circle graph.Express each result to the nearest cent. (Change percents to eitherfractions or decimals.)

How the Allen Famil.y Budgets

its $18,750 Annual Income

Housing

30%

Trans-portation

10%Taxes

22%Savings

2%

11. Which budget item is N cc the cost of housing?

Taxes are twice as much as which item?

3. What is the total percent spent on food and housing?

4. Find the dollar amount budgeted for each item.

5. How much more is spent than is saved?

SUMMARY:

What are the two methods for finding a percent of a number? How do youdecide which to use? Why?

60 65

LESSON 8

AIM: To locate points on a number line, to find the distance between twopoints and to locate a point midway between two other points


- locate any signed number on the number line.

- state that the number associated with a point on the number line is

called its coordinate.

- locate a point midway between two other points and determine its

coordinate.

- find the distance between two given points he number line.

- state that the midpoint of a line segment is the average of thecoordinates of the end points.

VOCABULARY:

New terms: coordinate, midpoint

Review: average, positive, negative, distance, number line,

opposites, additive inverses

CHALLENGE PROBLEM:

In a movie theater, there are two sections of seats, one to the left ofthe center aisle and one to the right. The theater is already dark

because the show has begun. The usher tells you that there is an emptyseat 5 seats from the center in a certain row. Does this informationtell you exactly where to look for the empty seat?

DEVELOPMENT:

1. Elicit from the pupils the fact that, since the usher did not saywhether the seat was to the right or to the left of the centeraisle, the exact location of the seat is uncertain. Pupils should

note that the seat could be in either of two places. Translate

these thoughts into a geometric model, a graph in one dimension(a number line). Draw a number line on the board, as shown:

< -lb -6 -43 j7 26 25 -4 -.2 21 b 1 3 S t 3 !3 io

Using the center aisle to separate seat numbers considered positivefrom those considered negative, elicit the following coentionsfrom the class: the center aisle location may be represented asthe "0" point :in the number line, all seats to the right of thecenter are designated with positive integers, and all seats to theleft of center are designated with negative integers.

2. Ask the class how the two seats that are possibly empty might bedescribed in terms of the number line. The pupils should readily

respond with the answers "+5" (or "5") and "-5". Ask the class tolocate on the number line a variety of seats to the left (negative)

C6

61

and to the right (positive) of the center aisle. Also mark severalpoints on the number line and ask pupils to read their coordinates.For this exercise, use a number line marked only with the zeropoint:

Define the number naming the 'Dcation of a point on a line as thecoordinate of that point.

3. Ask the class how far apart are the trio seats discussed. They mayobserve that the answer is 10, since each seat is S spaces to theright or left of the center aisle. This question is equivalent toasking for the distance between -5 and +5. Refer to the numberline on which +5 and -5 have been indicated. Lead pupils to deter-mine this answer by counting the number c space- between -5 and +5.


5. Call to the attention of the class the fact that the "center aisle"in the theater is given that name because is is midway between theseats on the right and the corresponding seats on the left. Thegeometric equivalent of this observation is the fact that the zeropoint on the number line is the midpoint of the line segment join-ing any two points whose coordinates are +a and -a. Illustratethis concept with such points as those with coordinates +5 and -5,+7 and -7, +3 and -3, etc. (These number pairs a.e also calledopposites or additive inverses.)

6. Generalize to points other than those with coordinates of +a and-a in the following manner: Have pupils locate points on thenumber line midway between any two other points on the line withintegral coordinates. Permit them to do this by counting. Forexample, to find the point midwa;. between +5 and -3, the pupilsshould note that these points are 8 units apart on the number lineand that the point midway between them is 4 units distant from eachof the points. Recall that movement to the right on a number lineis positive and movement to the left is negative. To get to themidpoint of the line segment joining the two points, it is necessaryto move 4 units to the right of -3 3+ (+4) = + 11 or move 4 unitsto the left of +5 r..1.5 (-4)= + 1 . Therefore + 1 is the midpointof the line segment joining + S an - 3, or +1 is the number halfwaybetween + S and - 3.

7. Elicit that finding the midpoint of a line segment is equivalentto finding the average of the coordinates of the endpoints. Recallthat the average of two numbers is found by adding the numbers anddividing by 2. Thug the average of +5 and -3 is:

(+ 5)+ (- 3) +22 2


APPLICATIONS:

A. For each of the following pairs of points locate the points on anumber line and find the distance between them:

6762

1) + 1 and + 9 5) - 7 and -2

2) 0 and +3 6) -1 and + 4

3) 4 and 10 7) - 2 and + 10

4) - 3 and - 9 8) 6 and 10

B. For each of the coordinates listed in Application A, find the coordi-nate of the midpoint of the line segment joining the points, bycounting. Check your results by averaging each pair of numbers.

C. 1. In a theater, you are sitting in the third seat to the right of thecenter aisle and your friend is sitting in the ninth seat to theright of the center aisle in the same row.

a) Show these facts on a number line.

b) How many seats apart are you and your friend?

c) What is the number of the seat which is halfway betweenyour friend's seat and your seat?

2. In the auditorium, two pupils are sitting in the same row. One pupilis sitting seven seats to the right of the center aisle while theother is sitting three seats to the left of the center aisle.

a) Show these facts on a number line.

b) How many seats apart are these two seats?

c) Which seat is midway between these two seats?

SUMMARY:

1. Review the fact that points on the number line have coordinateswhich are positive, zero, or negative, according to their locationon the line.

2. Also review the method of finding the coordinate of a point midwaybetween any two other points on the number line and the method offinding the distance between any two points on the number line.

CLIFF-HANGER:

If there were only one empty seat in an entire dark movie theater, what

information would you nee from the usher in order to find that seat?

63

LESSON 9

AIM: To name points on the coordinate plane using ordered number pairs

PERFORMANCE OBJECTIVES. Students will be able to:

- describe a location in a ,:oord'iate plane using two numbers (coordil _tes)-

- distinguish between location (3, 2) and location (2, 3), and definean urdered number pair.

- identify coordinates of a point to the left of 0 on the horizontalnumber line as negative, and similarly, coordinates below 0 on thevertical number line as negative.

label the horizontal number line with x, and call it the x-axis(horizontal axis).

- label the vertic number line with y, and call it the y-axis (- erticalaxis).

- locate the origin and state i,s coordinates.

- read and locate the coordinates of given points on the coordinateplane, including points on the axes.

VOCABULARY:

"ew terms:, x-axis, y-axis, coordina' axes, origin, quadrant, orderedpair, coordinates of a po

Review: coordinate

CHALLENGE PROBLEM:

A theater has a wide aims ",3 dividing the front and rear sections as wellas a center aisle dividiog the right and left sections of the theater.How many sections does this _heater have?

DEVELOPMENT:

1. Using the challenge problem, elicit from the students that thetheater is divided into four 5-c-c.lons and can be represented ina diagram as follows:

Left frontsection

Right frontsection

Wide aisle

Left rearsection

s4

0

a0U

64

6J

R It gar

sect ..,n

Using the aisles as reference lines, ask students what informatthey vould need to tind a particular location in the theater.Elicit that two pieces of information are all that is needed tofind any location (in this theater). The riPc?f, of informationneeded are the row number and the seat number.

2 Develop the analogy between aisles and coordinate axt._. Label

them x and y and refer to them as the "x axis" and the"yaxis".The sections of seats correspond to the four quadrants in the plane.Mark integral values alc 1g both axes, in a manner similar to thatused on the number lire it the previous lesson. Explain that,just as points to the right of zero were designated as positiveand those to the left of zero were designated as negative in theprevious lesson, so we now also designate points above zero (onthe vertical axis) as positive and those below zero as negative.Elicit the fact that the y axis is another number line. The zeropoint on the plane may be referred to ar the "origin" (the "original"reference point from which measurements are made in the horizontaland vertical directions). The following diagram should now be onthe board: Compare this diagram to the diagram of the theater.

Indicate that the axes form right angles at the origin.

In comparing the two diagrams consider such questions as:

a) What quadrant corresponds to the left front section?

b) Which aislt corresponds to the x-axis?

c) Where in the theater is the origin located?

3. Uz.,i.ng the diagram of the coordinate plane, kfigure 2) choose a

location labeled E in the diagram. Have students state how theywould describe the location of point E if the origin is used asthe starting point. Students may respond: go 5 units to theright and 3 units up,or 3 units up and 5 units to theright. Inform the students that the usual order in which thisinformation is given is 5 units to the right and 3 units up.This can be simplified by writing the ordered pair of coordinates

(5, 3).

Definition: In an ordered pair of coordinates, movement to theright or left of the vertical axis is indicated first. The secondnumber indicates movement up or down from the horizontal axis.These coordinates are written within parentheses and separated bya comma.

What are the coordinates of point F in figure 2?students observe that the coordinates (5, 3) namepoint than the coordinates (3, 5). Emphasize thement right or left of the y-axis is always namedordered pair.

4. Do Applications.

APPLICATIONS:

Figure 3

YA.

(3, 5) Havea differentfact that move-

first in an

7

NP-4 2 -7 1 -s

x

-y

5. U r7

T.

66

V

7-&

71

Use Figure 3 to answer all of Part A.

A. 1. Give the coordinates for ..:2ch of the points show(in Figure 3).

2. In what quadrant is each of the following points located?

B.

(a) B

(b) M

(c) S

(d) Y

(e) R

(f) C

3. Connect each set of alphabetically lettered points. What figures

do you get?

Set ABC:

Set KLMN:

Set RSTU:

Set WXYZ:

4

3

2

4-1-4...-4- 41-0-5 -4 -3 .2 -I i I 2 3 4 5I."MUM

1. How many units is P from the vertical axis?

2. How many units is P from the horizontal axis?

3. Give the coordinates of the following points:

(a) The point 4 units below P.

(b) The point 4 units above P.

(c) The point 4 units to the right of P?

(d) The point 4 units to the left of P?

(e) The point 3 units to the left of P and 2 units Mow it.

SUMMARY:

Review the manner in which the coordinate plane is divided into four

quadrants by the coordinate axes and how the locations of points in

the plan" are described by ordered number pairs.. Identify coordinates of

the origin. Review the new vocabulary introduced in this lesson.

67

LESSON 10

AIM: To locate points on the coordinate plane


plot points on the coordinate plane, given their coordinates.

state that the first number in an ordered number pair is called thex-coordinate and the second is called the y-coordinate.

VOCUULARY:

New terms:

Review:

x-coordinate, y-coordinate, plot

coordinate plane, ordered number pair, coordinate axes,quadrant, coordinates of a point, origin

CHALLENGE PROBLEM:

Lorraine leaves her house at (3,6) ar- heads for the drugstore which islocated at (-2, -8). However she me :s Rochelle at (-1, -5) and theythen decide to go to the candy store at k-3, -2). Will they pass thedrugstore on their way to the candy store?

DEVELOPMENT:

1. Refer to the challenge problem. Have students determine that, inorder to answer this question, a diagram of the coordinate planemust be drawn. Elicit the following:

(a) A set rf coordinate axes must be drawn and labeled in theappropriate manner. (See diagram below.) Emphasize that thescale on each axis is uniform.

14;

-1 -.7 ; -Of -J

a

ra )1x

r1

Pc

00.

68

73

(b) Review the meaning of ordered pair. Introduce the phrase"plot a point" as locating an ordered pair in the coordinate

plane. For example (3,6) indicates a point 3 units to theright of the vertical axis and 6 units above the horizontalaxis. 3 is called the x-coordinate of the point (3,6) and 6is called the y-coordinate of the point (3,6).

What information is given by the x-coordinate of an ordered

pair? (distance right or left of the vertical axis)

Complete the solution to the challenge problem.

2. Do applications.

APPLICATIONS: Note to teacher: One set of coordinate axes should be usedfor Application B (all parts) and one set for Application D

(all parts).

A.

13

N T

A

4

.t students find the answer to the .ciddle: "Why did the student plant

birdseed?" by following the directions below. Each point on the graph

is lettered. To answer the riddle, write each letter above the matching

ordered pair.

(2, -3) (2,0) (4, 3) (0, 4) (-2, 2) (1, 3) (-1, S) (-3, 4)

(4, -1) (-1, 1) (3, 0) (-1, -1) (-3,-1) (1, 4) (6, 1)

(1, -2) ( -4, 2) (-3, 3) (-2, -3) (0, -2) (4, 1) (0, -3) (0, 2)

69

B. 1. Connect the following points in alphabetical order. Then connectpoint M to point A:

A (10, 10) F (-8, 0) J (-3, -4)

B (8, 4) G (-6, -6) K (4, -6)

C (-10, 8) H (-8, -8) L (3, -8)

D (-16, 2) 1 (-8, -5) M (6, -8)

E (-12, -2)2. Shade the quadrilateral whose vertices are: (or whose corners ere

the points:)

(-10, 8), (-10, 3), (-8, 3), (-5, 7)

3. Shade the triangle hounded by:

(-16, 2), (-15, 1), (-15, 3)

4. Make a large dot at (-12. 4)

5. Name your figure.

C. Connect the following points in order and name the figure formed:

A (1, 2) G (-4, 0) M (1, -11)

B (6, 15) H (-4, -3) N (2, -8)

C (8, 18) I (-7, -6) 0 (2, -4)

D (6, 18) J (-7, -9) P (3, -2)

E (0, 2) K (-6, -11) Q (3, 1)

F (-2, 2) L (-2, -12) R (1,,2)

D. 1. Connect the fo'lowing points in alphabetical order.U to A.

1A (6, 14) I (-4, 25) Q (3,1B (2, 10) J 2) R (4,

C (2, 3) K (-7, -1) S (4,

D (-5, 10) L (-6, -3) T (3,

E (-9, 13) M (-3, -5) U (6,

F (-8, 4) N (-3, -7)

G (-6, 6) 0 (-1, -6)1

H (-1, 32) P (-1, -7)

Then connect

-6)

-3)

1)

3)

6)

2. In each set of coordinates connect the points in order:

a. (-5, -7), (-5, -91), (-2, -7), ( 2, -10), (-5, -7)2

b. (-1, -8), (3, -7), (3, -8), (-1, -9), (-1, -8)1 1 1 1c.

10), (-22-, 12), (-42-, lf), (-42, 0), (-2-T, 0)

70I 1J

SUMMARY:

Review the technique of locating points on a plane given their coordinatesand determining tie ordered number pairs associated with given points.

CLIFF-HANGER:

Plot R(-2, -2), S(-2, 4) and T(6, -2)

Find the perimeter of the figurt. formed.

71

rAIM: To find the distance between 2 points on the coordinate plane

LESSON 11


- find the length of a line segment that is parallel to one of thecoordinate axes.

- apply the Pythagorean Formula to find the length of a line segment thatis not parallel to either coordinate axis.

- find the distance between 2 points on the coordinate plane.

VOCABULARY:

Review: Pythagorean Formula, coordinate axes, perimeter, right triangle

CHALLENGE PRO: LEM:

Given: C 0), B (0, 4), A (3, 0)

Plot these points and connect these points to form a triangle.

What is the length of AC? of BC?

DEVELOPMENT:

1. Discuss the challenge problem. Recall that both he x-axis and they-axis are number lines. Review the procedure for finding thedistance between 2 points on a number line by counting (countingthe number of unit spaces between them).

Elicit BC = 4 units and AC = 3 units.

2. Refer to the cliff-hanger problem from the previous lesson.Develop technique for finding the lengths of line segmentsparallel to either AXIS. Elicit that the method of counting thespaces between the end points may be used to find the length of aline segment which is parallel to either axis.

Therefore:

Fig. 1

RT = 8 units

RS = 6 units


4. Return to the challenge problem. Ask students to find the lengthof AB. Why can't the answer be 4?

72

77

Elicit that this problem can't be solved by the method of countingsince there are no longer unit spaces along the line. One methodof obtaining the length of AB is to use another piece of graphpaper as a ruler. Place this piece of graph paper as shown below:

Fig. 2 Determine that thelength of AB = 5 units

5. A more direct method of finding this result should be developedas follows:

a) What type of triangle is aa, ABC? (right triangle)

b) What name is given to side BC and AC? (legs)

c) What name is given to AB? (hypotenuse)

d) What formula can be used to find the length of the hypotenuseof a right triangle when its legs are known? (PythagoreanFormula) Show how the Pythagorean Formula may be used to findthe length of AB.

(leg)2+ (leg)

2 = (hypoter)2

Therefore

or2 2 2

a + b" = c

42

+ 32

= c2

16 + 9 = c2

25 = c2

5 = c

Apply the Phythagorean Formula to find the length of ST in thecliff-hanger problem.

Since leg RT = 8 and leg RS = 6,

8-/+ 6

2= (ST)

2

64 + 36 = (ST)2

100 = (ST)2

10 = ST

Complete the solution to the cliff-hanger problem reviewing the

definition of perimeter.

73 P78

6 Apply the Pythagorean Formula to Pld the distance between any 2points in the coordinate plane.

Pose the Problem: What is the distance between (1, 1) and (6, 13)?

Elicit the following procedure:

a) plot the points and draw the line segment between them.

b) construct a right triangle with this segment as the hypotenuseand the legs parallel to the axes. See Figure 3.

c) find the length of each leg.1-J1

d) use the Pythagorean Formula tofind the length of the hypotenuse(distance between the 2 givenpoints.

d2= 12

2+ 5

2

d2= 144 + 25

d2= 169

d = 17

APPLICATIONS:

A. Draw each of the following line segments and find its length

1. (2, 1) and (9, 1)

2. (2, -2) and (2, -7)

3. (6, -2) and (-1, -2)

4. (-1, -3) and (-1, 5)

B. Find the distance between each pair of points.

1. (2, 1) and (5, 5)

2. (-8, 1) and (-3, 13)

3. (2, -3) and (-4, -11)

4. (-2, 3) and (4, -5)

S. (1, 4) and (6, -1)

ts)f133

[103

[10 3

ir5153

C. Find the perimeter of each of the following quadrilaterals:

1. A (4, u), B (20, 0), C (8, 5), D (4, 5) TrareLoid

2. A (0, 2), B (4.5), C (4, -4), D (0, -7) Parallelograr

SUMMARY:

Review the procedures for finding the distance between 2 points in theplane.

74

LESSON 12

AIM: To determine the coordinates of the midpoint of a line segment


- state that the x-coordinate of the midpoint of a line segment is theaverage of the x-coordinates of the endpoints.

- state that the y-coordinate of the midpoint of i line segment is theaverage of the y-coordinates of the endpoints.

- determine the coordinates of the midpoint of a line segment.

VOCABULARY:

Review: x-coordinate, y-coordinate, midpoint, average

CHALLEN.: PROBLEM:

When plotted on the same set of coordinate axes, Ship A is locatedat (1, 4) and Ship B is located at (7, 12). If the ships travel towardseach other at the same rate of speed, at what point will they meet?

DEVELOPMENT:

Figure 1

4

8

2

GO

41

8 9

1. Discuss the challenge problem using a graph of the given information(see figure 1). Elicit from students that the solution to theproblem involves finding the midpoint of the line segment joiningpoints A and B. Recall the procedure for finding the midpoint ofa line segment on a number line (a coordinate axis). Recall that

75 Li 0

to find the midpoint we averaged the endpoints. Since each pointin the coordinate plane is named by a pair of numbers (x-coordinateand y-coordinate), it seems logical to try to find the coordinatesof the midpoint by averaging the x-coordinates and by averagingthe y coordinates as shown:

=x +xx at the midpoint at A at B

2

= 1 + 7

2

8

2

4

=y +y

y at the midpointat A at B

2

= 4 + 122

= 16

2

= 8

Therefore the coordinates of the midpoint are (4, 8).

2. How can it be verified that (4,8) is the midpoint of AB? Rec 11that a midpoint is located the same distance from each endpoint.Have students find the distance from the midpoint to each endpoint(by using the Pythagorean Formula). See Figure 2.

tutPI,

-13

BO, z)

44-

ID

Le,

S7

f;

2

-/1

3

41ma.

76

b -8

61

(Am)2 32 42 (mB)2 32 42

(AM) 2 = 9 + 16 (MB) 2 = 9 + 16

(AM) 2 = 25 (MB) 2 = 25

= 5 MB = 5

Therefore, AM = MB which shows that M is the midpoint of AB.

3. Do applications.

APPLICATIONS:

1. Find the coordinates of the midpoint of each of the line segmentswhose endpoints have the coordinates shown:

a)

b)

(3, 5) and (9, 13)

(-2, -6) and (-10, -8)

c) (1, 4) and (7, -4)

d) (5, 3) and (-9, -1)

e) (0, 0) and (6, 6)

f) (-9, 0) and (12, 0)

g) (1, 3) and (3, 1)

h) (-6, 0) and (0, 10)

2. Have students plot each line segment used in example 1 and showthat the midpoint is the same distance from each endpoint.

SUMMARY:

Review the method for finding the coordinates of the midpoint of a line

s..jment having students state that the x-coordinate of the midpoint isthe average of the x-coordinates and that the y-coordinate of the mid-point is the average of the y-coordinates.

77

UNIT IX. Ban king and TaxesLESSON 1

AIM: To compute simple interest

PERFORMANC7 OBJECTIVES: Students will be able to:

- use the words deposit, withdrawal, principal, interest, interest rate,and balance to describe banking procedures.

- explain why 1. bank rays interest.

- state that interest rates are always expressed as an annual rate.

- identify the principal, rate, and time period in a bank interest. problem.

- compute interest using the simple Llterest for; I = Prt, where tis an integral number of years.

VOCAFULARY:

New terms: principal, interest, intere:t rate, balance, annual rate

Review: :ithdr-

CHALLENGE PROBLEM:

Mr. Rodriguez wed $10 ich week his new savings account. By theend of one year he had made 52 deposits and his bankbook showed a balanceof $539.80.

1. How much money did he deposit in 52 weeks?

2. Where did the extra money come irom?

DEVELOPMFNT:

1. Discus' the challenge proble..i. Elicit that the extra money is calledinterest. Define interest as a rental paid for the use of money.When money is deposited in a savings account, the money is used bythe bank. The bank then pays the depositor inter( -t (a rectal fee).Similarly, when someone borrows money (takes a loan), they must payInterest (a rental fee) to ne lender.

Examine bank advertisements and elicit the following points:(see figure 1)

a) Interest rate is expressed as a percent.

b) interest rates are given per yea_ or per annum.

c) Princlp-1 is the amount on whit}- 'aterest is figured(deposited or borrowed).

d) Deposit is money added to an account.

e) Withdrawal is money taken out of an account.

f) Balance is the amount of money in an accnv.nt at any given

79

63

Figure 1 11,Small money

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yieldon a 5-ttlonth

26lime Savings

;countis an ..annual

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lotveg. I% avallable

at tnatuty,26meek

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if the bankconsents;6E113 months'

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S

3. Pose the problem: If an interest rate is 9% per year, findthe amount of interest on a principal of $500 fcr o year.

a) What information is given? (interest rate, principal,length of time)

b) What information must be found?

c) Elicit that the soluticn means 9% of $500 must be found.

d) Obtain the solution by the following calcuation:

Interest = 90 of $500 Calculation

Interest = .09 x 500 $ 500

Interest $45.00x .09$45.00

e) If the interest rate remained the same, hob, much interestwould te paid in three years?

Elicit that the solution is found by multiplying thL yearlyinterest by 3. In cases where the interes is computed onthe same principal each time, the interest is calledsimple interest.

4. Have students state in words the procedure used to calculate thesimple interest. (Interest is found by multiplying the principalby tLe rate of interest by the number of years.)

Elicit the formula' I = Prt Jwhere I = interest, P = principal,r = rate of interest, t = number of years.

5. Do applications.

APPLICATIONS:

1. Find the interest on a $900 deposit for one year at 7%.

2. How much interest will Mr. Sargent receive if he deposits $1500for two years at 10% interest each year?

3. Nr. Lee makes an investment of $5000 that yields 15% interest. Henuch will the investment earn in one year?

4. A & B Auto Repairs borrowed $4000 to purchase new tools. If theloan was repaid at the end of 2 years and the rate of interest was21%, find the interest due.

SUMMARY:

Use the following problem to iwiew the major concepts in this lesson.

$12,000 was borrowed for a period of 3 years at an interest rate of 19%.

a) What is the principal?

b) What is the annual rate of interest?

c) What is the interest charged?

d) How much money must oc repaid at the end of three years?

81 6J

CLIFF-HANGER:

A savings bank pays 7 -% effective annual yield on a deposit of $500.

How much interest is earned in one year?

828 6

LESSON 2

AIM: To compute simple interest


- express non-integral percents as decimals.

- use the simp3e interest formula to calculate interest where the rateis a non-integral percent.

VOCABULARY:

Review: percent, interest, interest formula, interest rate

CHALLENGE PROBLEM:

A bank advertises 12.94% annual yield on deposits of $500 or more. Howmuch interest is earned in one year on a deposit of $600?

DEVELOPMENT:

1. a) What information is given in the Challenge Problem?(interest rate - 12.94%, principal - $600 time - 1 year)

b) What formula can be uses to solve the problem? (I = Prt)

c) Substitute the v lues in the formula to obtain the statement:

I = $600 x 12.S4% x 1

How must 12.94% be written so that the calculation can beperformed?

2. Deve:op a procedure for changing a percent containing a decimalinto a decimal number as follows:

a) Recall tne meaning of percent as the ratio of a number to 100.

b) Express 12.c4% as12.94

100

.1294

c) Divide 100 )12.9400-10 0

2 94

-2 00

940

-900

400

-400

Therefore 12.94% = .1294.

d) Recall the shortcut method for division by a power of 10. Since100 = 102, the result of division by 100 can be found by placingthe decimal point 2 places to the left of its original positionin the givim number. Demonstrate: 12.94% = .12 94.

e) Complete the solution to the challenge problem.

8387


4. a) Consider the cliff-hanger problem from the preceding lesion.

In order to solve this problem, the student must substitute0

the values $500, 7-s and one year into the simple interest

formula obtaining:

I = $500 x3

x I4

b) Using a procedure similar to the one outli.ed in item #1, elicit0

a need to express 7-4-6 as a decimal

5. Develop a procedure for changing a percent containing a fraction

into a decimal imiber as follows:

a) What is the difference between the form of the annual rate in

the challenge and cliff-hanger problems? (One contains

a fraction, the other contains a decimal.)

30b) Hew can 7-4-6 be expressed so that the procedure learned previously

can be used? Elicit that 7I-must be expressed as a decimal.

c) Review changing a fraction to a decimal by division:

4

3--means 4)3.00

.75

_2 5.1

2.0

-207 3

Since :=1: = .75, 7t-r= 7.75.

30d) Therefore 7-4-6 = 7.75%. Now the student can change 7.75% to

.0775 using the procedure previously learned.

6. Complete the solution to the cliff-hanger.


APPLICATIONS:

A. 1. Express each of the following percents as a decimal:

a) 14.48%

b) 17.5%

c) 15.301%

d) 7.5%

e) 8.03%

f) 7.25%

g) 5.75%

h) 9.051,96

2. the annual yield on a minimum deposit of $10,000 f '5.22%. Find

the interest receive-I. for two years on a deposit (_ $10,000.

84 68

B 1. Express each of the following percents as a decima.:10 0

a) 12v. e) 5,D

10 30b) 18v f) 84-i.

30 , ,30c) 174,1 g) 6-<,

8

So 10vd) 12-i h)8

2. A bank charges 14% interest on a credit card cash advance. If

a cash advance for $800 is paid off in a lump sum after one year,how much interest is charged?

SUMMARY:

Review the procedures for changing non-integral percents to decimals.

CLIFF-HANGER:

What would be the interest charged on aa 18% loan of $3000 for a periodof 15 months?

85tl 9

LESSON 3

AIM: To compute simple interest involving fractional parts of the year


- express any number of months as a fraction of a year.

- express a non-int gral percent as a fraction in lowest terms.

- use the simple interest formula to calculate interest where the rate is

a non-integral percent and the time is non-integral number of ye,?rs.

VOCPPULARY:

Review: annual, percent interest

CHALLENGE PROBLEM:

If $120 interest is earned in one year, at the same rate how much inter-

est is ea led after 6 months? after 3 months?

DEVELOPMENT:

1. Discuss the challenge problem. Have students recall that the rate

of interest on a deposit or loan is always an annual rate. Elicit

that 6 months or 3 months is a pa.t of a yea.. Have studentsdetermine the fractional part as follows:

There are 12 months, so each month is1

of a year. Therefore:

,

6 months =6 3

a year; 3 months = of a year.

1Reducing each fraction, 6 months =

2of a -ear and 3 months =

of a year.

Have students generalize that when a loan is made for a period otherthan a whole number of years, the time mist be expressed as amultiple or part of a year.

Solve challenge problems as follows:

interest earned after 6 months = of $1202

interest earned after 6 months =1

(120)2

interest earned after 6 months = $60

1interest earned after 3 months = -zr of $120

1interest earned after 3 months = zi (120)

interest earned after 3 months = $30

2. Do Applicatioi. A.

3. Consider the cliff-hanger problem. Elicit the simple interestformula (I = prt)

ISElicit 15 months =

12or years.

Guide students to discover that the calculation would be simpler ifthe interest rate were written as a fraction rather than as a decimal.Therefore: 18

18% =100

I = Prt

30I = $50efot x 18 x 5

*0379-

IS

18 5I= x x --4-1

2

915 3

I = x --+8- x1 1 --2--

I = $675


5. Present the problem:

At 720, what is the interest on an $800 loan for 1 year 9 months?

Elicit the steps nt-eded to solve his problem:

I = Prt

9t = 1 yr:- 9 months = 1 1-2- years

t = 13

years

7t = 71 years

Guide students to discover that the calculation would be simpleris 71/2% were expressed as a fraction as follows:

71; 71 71 1520 T or -2- .: 100 or - -100

100

1 IS 1007-4 =2 2 1

31

7 - =44-

x2 2 44n0

17-4 = --2 40

Complete the solution:

3 7

I

I = $800 x40

x74:

4204

I = Sins


APPLICATIONS:

A. Express i lowest terms each of the following, using fractional partsof a year:

1. 5 months 5. 1 year 4 months

2. 2 months E. 5 years 6 months

3. 4 months 7. 4 years 8 months

4. 13 months

B. 1. Marti.a made a $400 loaf! for 1 year ' months at 15%. Find theinterest she owed.

2. Mr. breg owed $900 in taxes for 20 morths. The penalty was a 6%interest charge. How much interest was he charged?

C. 1. Express each of the foll(wing ?arcents as a fraction:3 5

a. 5% d. 12T%4

1b. 12-% e. 33i%

2

1 3c. 18-% f, 6T%

4

2. A credit card account charges 171/2% interest on unpaid balances.Sandra owed $200 on her credit card. What was th,: interest charge

for one month?

S1 %4MARY:

Rev;ew changing percent to fractions and use of the simple interest form, a.

CLIFF-HANGER:

A bank advertises 6% interest compounded daily. What does this mear?

88

LESSON 4

AIM: To understand the meaning of compound interest


- differentiate between simple and compound interest by stating:

1. that when calculating simple interest:

a. the interest is figured all at once for the entire period,

b. the principal does not change during the period.

2. that when calculating compound interest:

a. the interest is figured separately for each unit of time(e.g., per year for annual compounding),

h. the principal incases after each unit of the time perio'(e.g., this year's principal equals last year's principalplus last year's interest).

calculate interest coypounded annually.

VOCABULARY:

New terms: compound interest, inter-st periods

CHALLANGE PROBLEM:

How much money will you ha:e in the bank at the end of 1 year if youueposit $500 at 6% interest?

DEVELOPMENT:

1. Discuss the challenge problem. Recall that I = Prt is used tocompute simple interest. What happens to the $30 interest thatthe bank pays? (It is credited to the account.) Therefore, thesolution to the challenge problem is $530.

2 Now pose th,- problem: Suppose the money is left in the accountfor anothe year. On what amount of money (principal) will thebank be paying interest? Elicit the response $530. Using thisas the principal, compute the interest paid for the second year at

same rate of interest. Then compute the 1- '.oce:

I = Pit

I = $530 x .06 x 1

I = $31.80

Balance = previous balance + interest.

Balance = $530 + $3i.80

Balance = $561.80

3 Have students compute the simple interest for $500 at 6% for 2 years,and the Final balance.

890'4t1

I = Prt

I = $500 x .06 x 2

I = $60

Balance = $500 + $60

Balance = $560

4. Compare the results of both computations. Have students speculatewhy the balances are different. Discuss with the class the factthat the procedure follow J in item 2 is called compounding inter-est. In compounding interest, as interest is earned, it is addedto the principal. This creates a new principal, on which the newinterest is calculated.

5. Discuss the fact that many banks compound interest quarterly. Whatdoes this mean? (Interest is paid every 3 months; in other wurds:

1_ )12 4

Have students recognize that4

year represents the time, t, used

in computing the nterest for the 3 month period.

6. Recall the cliff-hanger problem. What value oft is used in com-

puting the daily interest in a normal year? (-37

) Computations

involving compounding are generally done using a calculator or

tables.

7. Do applications.

APPLICATIONS:

!. What is the total interest earned if $2000 is deposited for 2 years

at 15% compounded annually?

2. Mr. Ftrong :leposited $4,000 at 8% compounded annually.

a. How much will he have on deposit at the end of two years?

b. How much will he have on deposit at the end of three years?

3. Have pupils find examples of sums of mono/ deposited in banks

which have grown to large al nts due to compounded interest.

SUMMARY:

Have pupils state that:

1. when calculating simple interest- -

a. the interest is figured an at once for the entire period,

b. the principal does not charge during the period.

2. when calculating compound Interest- -

a. the interest is figured separately for each unit of time

te.g., per year for annual compounding),

b. the principal increases after each unit of the time period

(e.g., this year's principal equals last year's principal plus

last year's interest).

90 `0, 4

LESSON 5

AIM: To develop the concept of equal repayments for a loan


- calculate the amount or interest charged on a loan.

- calculate the total amount to be repaid (interest plus principal).

state that each installment is the quotient of the total amount borroweddivided by the number of equal installments.

- compute each installment on a loan.

VOCABULARY:

Review: interest, principal, installment, mortgage

CHALLENGE PROBLEM:

Mr. Williams borrowed $1000 for 2 years at 10%. How much did he haveto repay?

DEVELOPMENT:

1. Have students recall that a borrower must repay the full amountborrowed, called the principal, plus the interest owed. Let the

class calculate the amount of interest dues

I = Prt

I= icoo x10 x2100

I = 200

Mr. Williams must repay $1,000 plus $200 interest, for a total o$1200.

2. Mr. Williams plans to repay this loan in equal monthly installmentsover the 2-year period. How much will he repay each month?

Guide students to an understanding that there will be a total of 24

monthly payments in a two-year period. Since the payments are to be

equal, $1200 i 24 = $50, Mr. Williams will have to repay $50 per month.

3. Do Application 3 (use this opportunity to reinforce calculationsinvolving a variety of percents).

APPLICATIONS:

1. Mrs. Quinn borrowed $1800 for 2 years at 12%. What was the amount

of each equal monthly repayment?

2. To purchase a house, the Hernandez family obtained a 10 yearmortgage for $40,000 to be repaid in equal monthly payments at 9%interest. What would their monthly mortgage payment be?

3. To pay for his daughter's wedding, Mr. Weber borrowed $900 to berepaid in 18 equal monthly installments. At 14% interest, what

was his monthly payment?

91

4. In order to buy a sports car, Fred borrowed $9000. He agreed tomake eval monthly payments over a five year period, at 11.5%annual interest. How much did he have tc repay each month?

5. The B & G Sportwear Corp. borrowed $18,000 at 141/2% to be repaid inequal quarterly (four times a year) payments over three /ears.Find the amount due each quarter.

SUMMARY:

Review the idea that loans are often repaid in equal installments andthat these payments include both the amount borrowed (principal) and theinterest owed.

D6

92

LESSON 6

AIM: To introduce the idea of "bank discount"


- -ompute the amount of interest charged on a loan.

- compute the actual amount of money available to the borrower (net

proceeds).

- determine whether a given amount yields sufficient net proceeds.

VOCABULARY:

New terms: net proceeds, bank discount

CHALLENGE PROBLEM:

The Second National Bank agreed to '..nd Mr. Freeze $500 for 6 months a'

8%. When Mr. Freeze went to get the money, he was given only $480.What happened to the other $20?

DEVELOPMENT:

1. Ask pupils to try to solve the challenge problem. If they fail

to do so, suggest that we consider one of the most importantaspects of any loan, the interest.

I = Prt

I = 500 x x100 12

I = 20

Ihe interest on Mr. Freeze's lore will be $20.

2. Explain that a bank often "discounts" or subtracts the amount ofinterest on he loan in advance. Thus, Mr. Freeze will have torepay $500 which includes the $480 he received,called the netproceeds, plus the $20 interest. (Net proceeds = P I)

3. Have the class consider this situation: Marlene wishes to purchase

a new car that will cost $4200. Her bank agrees to give her a

three year loan, discounte3 at 9%.

a. If she borrows $4200, will she receive enough money to buy the

car?

b. If she borrows $5000,wiil she have enough?

c. If she borrows $6000, how much more than the price of the gar

will she receive?

Solution to a.

I = Prt42

I = 4200 x2- x

-irae 1

I = 1134

93

Net proceeds = P I

Net proceeds = 4200 1134

Net proceeds - $3066

The answer is that she will not have enough money if sheborrows $4000.

Similarly for b.

I = $1350

Net proceeds = $3650.

The answer is that she will not have enough mcaey if she

borrows $5000.

Similarl for c.

I = $1620

Net proceeds = $4380

The answer is that $4380 is $180 more than the $4200 she needs.

4. Do applications.

APPLICATIONS:

1. Mr. Cohen borrows $1000 for 1 year discounted at 10%. How much

will he actually receive?

2. Martha borrowed $750 from her bank. If the loan was discountedat 8% for a period of six months, how much did she receive?

3. In order to purchase a car costing $4500, Mr. Walters borrowed$5000 for 2 years. The 10% interest was discounted in advance.

a. How much did he receive?

b. How much more (or less) than he needed was this?

4. The Mulligan Family want to two banks to discuss a loan. The

Northwest Bank offered them a $700 loan for 3 years at 12% simple.

interest. The Citywide Bank offered them a $1000 loan for 3 years

at 10%, discounted in advance.

a. Find the interest charge for each loan.

b. Which would have been the best choice? Why?

SUMMARY:

Review the idea tha4 interest is often deducted in advance ("discounted")on a loan, so that the borrower does not receive the fill amount of theloan.

CLIFF-HANGER:

In what situations other than banking is the term "discount" used? (Note

to teacher: Have students bring in newspaper advertisements showingdiscounts ana discounted prices.)

94 (3,-.5

LESSON 7

AIM: To compute the selling price when discounting


- compute the amount of a discount if the discount rate is a unit

fraction.

compute the selling price (net price), if the discount rate is a unit

fraction.

VOCABULARY:

New terms: discount, markdown, rate of discount, original price, saleprice, discount price, net price

CHALLENGE PROBLEM:1

During an "end-of-season" sale, a store gave a discount of off every

ski jacket. Mark purchased a jacket with an original price of $42. What

was the discount? How much did he have to pay?

DEVELOPMENT:

1. Use the challenge problem to list and discuss the meanings of the

words commonly associated with "sales". What do each of the

numbers stated in this problem represent? Elicit:

a. $42 represents the original price (also called the retailprice, regular price, ticket price, marked price or list price).

1b. . 3. represents the rate of discount.

2. a. Solve the challenge problem. Define the term discount (ormarkdown) as the reduction in the original price of the article.Since the discount was given as a rate, the dollar value of the

discount m be computed as follows:

Discount =1of original price

1 4.2.14

Discount = f x 7Discount = $14

Emphasize that the discount is the actual dollar amount of thereduction ($14) as compared to the rate of discount (4).

b. Have students compute the cost of the ski jacket as follows:

Sale price = ORIGINAL PRICE - DISCOUNT

Sale price = $42 - $14

Sale price = $28

Discuss other terms meaning "sale price" (discounted price, net

price)

95 9 9


4. Refer to newspaper ads:

Use the following example (or similar ones brought in by students)to determine the validity of ads.

RECORD ALBUMS - Regularly $9.99

Now1off - only $6.89

Can we assume that the sale price in the ad is correct?

Have the students compute the discount and the sale price. Havethem compare their sale price to the advertised sale price.


APPLICATIONS:

1A. 1. At -4- off, what is the discount you will receive when purchasing

a $64 coat?

12. Arlene purchases a $27 calculator during a Toff sale. What is

the markdown? What is the sale price?

3. Original Price: $12.991

Now: y off

Net price: ?

14. A newspaper advertisement read: Take Toff the price of all summer

clothes. Carl saw a $25 jacket he wished to buy. If he 11,:d $20,

was he able to make the purchase? (Disregard sales tax.)

5. Two department stores advertised the same TV game this way:

THE 'BARGAIN STORE

Our Regular Price: $72

This Week Only: off

GAMELAND

Original Price: $81

1Now: y off

Which store had the lower selling price?

B. The following items were placed on sale in a local store:

(a) (b) (c) (d)

B & W Radio Tape Color TVTV Recorder

List Price $128 $57 $89 $499

Discount1

4- off1T off

1T to1y off

1Tto 1i off

Sale $96 $38 $62 $399

Is the store correct in its advertisements?

,

C. Mrs. Cruz purchased a $90 rat for $60. How much of a discount had she

received? What fraction of the original price is the discount

(rate of discount)?

Note: Teacher may wish to determine the rate of discount for eachitem in Application B using the given list price and sale price.

SUMMARY:

R call that a discount is an amount deducted from the original price ofan article, and the rate of discount is often expressed as a fraction

of the original price.

Compare a store discount with a bank discount. Compare net price of a

purchase with net p.oceeds from a loan.

CLIFF-HANGER:

Employees of the M & V Stationery Co. receive a 15% discount on all

purchases. What will an employee pay a $39 file cabinet?

97101

LESSON 8

AIM: To find the selling price when the rate of discount is expressed as apercent


- convert several common percents to their fractional equivalents(10%, 20%, 25%, 33 1/3%, etc.)

- compute the net price where the discount rate is given as a percentby expressing the rate as a decimal or common fraction.

VOCABULARY:

Review: rate of discount, discount, sale price, etc.

CHALLENGE PROBLEM:

Employees at Marsh's Shoe Emporium receive a 20% discount on all purchases.How much of a discount will Maria, an employee, receive on a $50 pair ofshoes?

DEVELOPMENT:

1. Discuss the challenge and cliff- hanger problems. In what waydo tb'se problems differ from the problems in the previous lesson?(The rate of discount is expressed as a percent.) Recall variousnewspaper ads.

2. Review the procedure for finding a discounted price when the rateof discount is a fraction. Recall that every percent can be ex-pressed as either a fraction or a decimal.

20 1For example: 20% =

rub T or 200 = .20

3. Elicit that the challenge problem can be solved in either of twoways:

Discount = 20% of $50

a. 20% =1

Discount =1of $50

1Discount = -§ x

50

b. 20% = .20

Discount = .20 x $50

Discount = $10.00

Discount = $10

4. Elicit the two methods of solving the cliff-hanger problem.

The discount = 15% of $39

15a. 15% =

100b. 15% = .15

9811)2

15Discount = --

100of S39

Discount = 15 x 39100 1

Discount = 585100

Discount - $5.85

Discount = .15 Y $39

Calculation

$39x .151.95

3.9$5.85

Discount - $5.85

Compute Net (employee's) Price to be $33.15.

Note: Allow students to choose either method of solution. You

may wish to discuss with students their reasons for favoringeither method.

S. Do applications.

APPLICATIONS:

1. Find the discount on a $350 TV set being sold at 20% off.

2. A $40 pair of shoes is priced at 30% off.

a. What is the markdown?

b. What is the sale price of the shoes?

3. Regular price: $60. Now: 331% offlo Find the discount price.

(Nyte: SincT the decimal form rof 33 is not convenient for use,

express 331%- in fractional form.)

4. Members of the Art Guild receive a 5% discount on all purchases.

How much will Rosa save on a purchase of $26?

5. G.O. members can often attend 'pro' games at 50°4 off the normal

admission price. If the normal price is $4.50 per ticket, what

will it cost for a G.O. member?

6. For a Valentine's Day gift, Hector had to decide between a $30

necklace 20% off and a $40 bracelet at 40% off. Which would

have been the least expensive? (Have pupils realize that thelarger percent discount or the smaller original price do not, by

themselves, guarantee the cheapest selling price.)

7. At 20% off, find the cost of a $36 skate board.

8. For their 23rd Anniversary Sale, the G?obe Sporting Goods store

gave a discount of 23% off every item. If Bill buys an $18 basket-

ball during this sale, what will the sale price be?

SUMMARY:

1. Have students state that when the rate of discount is expres:,d asa percent, the percent must be given as a fraction or decimal

before computing the discount.

10399

2. Express some frequently used percents as fractions and decimals

(50%, 25%, 20%, 3314).

CLIFF-HANGER:

Carolyn purchased a cassette recorder marked $48.95. An 8% sales taxwas then added to this price. What was Carolyn's total cost?

LESSON 9

AIM: To find the total cost of a purchase


- calculate the sales tax (or tip) by finding the product of the

expediture and the rate of sales Ix (or tip).

- express monetary values to the nearest cent.

- comnute the total cost of a purchase including tax, tip, or both.

VOCABULARY:

New terms: Sales tax, tax rate, tip

Review: rounding off

CHALLENGE PROBLEM:

Paulette had dinner in a restaurant. The cost of the food was $12.50.

She told Linda that dinner cost her $15.50. How do you account for

this difference?

DEVELOPMENT:

1. Discuss the challenge problem. Elicit from the class, that sales

tax and a tip account for the $3 difference.

2. Elicit from students that sales tax is generally charged on most

items purchased. In some cases the sales tax is already includedin the cost listed (for example - gasoline, taxi rates, admission

tickets). In most cases the sales tax is a percent of the purchase

price and must be added on to find the total cost. The sales tax

ra'..e varies in different places and for different items.

Sometimes the amount of tax is determined by using a sales tax

table. When a table is not available or when the cost of thepurchase exceeds the scope of the table, the tax is computed by

using the sales tax rate (usually expressed as a percent).

3. In the challenge problem have students compute the sales tax if

the sales tax rate is 8%.

Solution: 8% = .08

Sales tax is 8% of $12.50

Sales tax = .08 x $12.50

Sales tax = $1,0000Recall that cents is expressed as a two plac' decimal. Therefore

$1,000 should be expressed as $1.00. Elicit from the class that

the tip Paulette left was $2.00.

4. Recall the challenge problem. The solution is as follows:

Find 8% of $48.95

$48.95

x .08

Sales tax = $3.9160

101105

Round off $3.9160 to $3.92. Since the total cost means the costincluding the sales tax, add $3.92 to $48.95. ($52.87 is thetotal cost.)

5. Have students state in words how the total cost of a purchase isobtained.


7. Return to discussion of the challenge problem. Elicit examples ofsituations where tips are generally given (services rendered).Explain that tips usually range from 15% to 20% of the cost ofservice before tax. These amounts are usually rounded off to thenearest 25t. Compute 15% of 812.50 and determine whether the tipwas adequate. Tip = 15% of $12.50:

$12.50x .1562 50125 0

$1.87 50 --- $1.88

Since $1.88 is an awkward amount to leave as a tip, Pauletterounded the amount of the tip up to $2.00. (Discuss why she left$2.00 instead of $1.75.)


APPLICATIONS:

A. 1. Arthur's new coat was priced at $77.50. At 8%, how much tax mustbe paid?

2. Carmen purchased a new bicycle, priced at $129. If the sales taxis 81/4%, how much tax must she pay? (Recall 81/4% = 8.25% = .082S)

3. Eileen's family saved $300 to purchase a new television set. Ifthe list price of the set is $289.50 and the tax is 8%, do they haveenough money to pay for the set? If not, how much more do theyneed?

4. Fred visited his uncle in Albany, where the sales tax was 5%. Ifhe bought two records at $4.75 each, how much did he have to pay?How much change did he receive if he paid with a $20 bill?

5. The tax on parking a car is 6%. After parking for 3 hours, Mr.Gordon owed $3.77 plus tax. What was the total he owed?

6. Bob saw a sports car priced at only $4550. How much sales tax, at71/4% must he add to the price?

B. Find a 15% tip for each of the following bills:

1. A haircut for $12.00.

2. A taxi ride costing $8.10.

3. A meal for $14.65.

SUMMARY:

Have pupils review the components of the total cost of a purchase.

CLIFF-HANGER:

Louis saw that a $25 calculator was on sale at 25% off. He knew that

an 8% sales tax would be added. If he had exactly $20, could he pur-

chase the calculator? How mu,:h more - or less - than the total cost

did he have?

103

LESSON 10

AIM: To calculate the total cost of a purchase involving both discountand sales tax


- state that sales tax is added after the sale price is calculated.

calculate the purchase price for aoods advertised in a newspaper saleincluding discounting and sales tax.

VOCABULARY:

Review terms: discount, sale price, total cost

CHALLENGE PROBLEM:

An item which lists for $25, sells for $20. If the tax rate is 8%,how much tax must be paid on this item?

DEVELOPMENT:

1. Elicit from pupils that sales tax is charged un the actual costof an item. If a discount is given, the sales tax is charged onthe reduced price of she article. Therefore in the challengeproblem the sales tax is 8% of $20 or $1.60.

2 Have pupils work out the cliff-hanger problem:

a) $25 original pricex.25 rate of discount

$6.25 discount

b) $25.00 original price- 6.25 discountT18.75 reduced price

c)$18.75 reduced price d) $18.75 reduced nricex .08 rate of tax +1.50 sales puce$ 1.50 sales tax $20.25 total cost

Louis could not purchase the calculator since he needed 25(t more.

3. Pupils may determine the total cost of actual items advertised inthe local newspapers. Thc' applications will be more meaningfulto the students if they relate to specific purchases of interestto the students.

APPLICATIONS:

(Note: In all the following problems, the teacher may specify thatthe effective local tax rate be used.)

1. Nina bought a bathing slit originally marked $25, for 40% off.Including the sales tax, what was her total cost?

2. During a1

off" sale, Bill purchased two shirts originally priced$7.50 each. Find his total cost when the sales tax was added.

3. The Aquarius Specialty Shop advertised: "25% off the firstsweater you buy - 50% off the second..." Edith purchased twosweaters, each with a list price of $18. Find her total cost.

1041n8

4. In Chuck's town, a $60 record player is on sale at 30% off plus

4% tax. Iy the big city nearby, he can purchase the same record

playcr at 3-cff, plus 8% tax. Where would he obtain the lower

price? Hnw much lower?

SUMMAaY:

Have the pupils state that the discount or reduction must be subtractedfrom the original price before the sales tax is computed.

109105

LP'"ON 11

AIM: To solve business and consumer problems involving rates


- state the meaning of common business terms involving the rate expressedas a percent (commission, markup, successive discount).

- compute commissions, markups, and successive discounts when the ratesare given.

VOCABULARY:

New terms: commission, markup, successive discount, across the board,base sticker price, surcharge, successive discount

CHALLENGE PROBLEM:

A salesperson receives a commission of 36% of total sales. How muchdoes the salesperson earn on sales totaling $642?

DEVELOPMENT:

1. Discuss meaning of commisAon. Commission is the money paid by anemployer, based on the value of the goods sold or the servicesrendered to an employee. The rate of commission is generallyexpressed as a percent. As with awl other problems where a rateis given, the commission is found by multiplying the rate ofcommission by the total value of goods, sales or services. Solvethe challenge problem as follows:

Commission = 36% of sales

Commission = .36 x 642

Commission = $231.12

2. There are many business situations involving rates expressed aspercents. Some of these are covered in the following applications.The teacher should supplement these with a variety of problemsemphasizing business terms and computation with percents.

APPLICATIONS:

1. In order to keep up with rising costs, an automobile manufacturerannounced a 9% across the board increase on the base stickerprice of a car. the price e P car was listed as $6390,

a. What is the increase in cost?

b. What is the new base sticker price?

c. If the automobile salesman makes a 12% commission, how muchmore will he receive after the price increase on this car?

2. Mr. and Mrs. Beale dine at their favorite restaurant. Mr. Bealeorders a steak dinner for $12.95. Mrs. Beale orders a lobsterdinner for $14.95. The menu states that there is a 10% surchargeon lobster.

a. Find the actual cost of the lobster.

b. Find the total bill for the 2 dinners if the sales tax is 8%.

c. If Mr. Beale leaves a 15% tip, how much change will he keepfrom a $50 bill?

3. By paying his bill within ten days, Mr. Fernandez receivedsuccessive discounts of 25% and 5%. If the items he purchasedhad a list price of $480, how much did he have to pay? (Explain

that to calculate successive discounts, use the first discountrate to find the first reduced price, and then use the seconddiscount rate on this reduced price to find the final reducedprice.)

4. Pat's father can purchase $15 footballs from a company that offersdouble discounts of 20% and 20%, or from a dealer who offers asingle discount of 35%. Which would be the least expensive?

5. A video cassette recorder costs a dealer $590. In order to coverhis expenses and make a profit, he must use a 35% markup. Atwhat price will he sell the recorder?

6. Ronnie's sister is paid a weekly salary of $150 and a 3% commis-sion on all sales over the first $1000 in sales each week. Duringthe week when her sales were $4297, what were her total earnings?

SUMMARY:

Review vocabulary of business and consumer problems involving rates.

107 111

UNIT X. AreasLESSON 1

AIM: To find the areas of rectangles


- provide practical examples where area is used (e.g., floor tiles,carpeting, wallpaper).

- state that a surface (area') must be measured with a square unit, andis different from perimeter which is measured with linear units.

- find areas of rectangles by counting the number of squares theyenclose on a grid.

VOCABULARY:

New terms: area, square unit, surface

Review: perimeter, linear unit, rectangle

CHALLENGE PROBLEM:

On a pair of coordinate axes plot the following points and draw thequadrilateral ABCD:

A(2,1), B(6,1), C(1,4), D(2,4),

a. What is the perimeter of quadrilateral ABCD?

b. What type of quatrilateral is formed?

DEVELOPMENT:

1. Using the challenge problem, have students recall the meaning ofperimeter (the number of linear units of measure around a figure)

and the method for finding the length of a line segment on a coor-dinate plane (counting the number of spaces on a line). The peri-meter is 14 units.

Figure 1

109

7

D Li

3

3 3

', - :. -vL.

2

/ 2 :.7.: ' l 7 s

-I--___ ..

" 2;

2. Elicit that the quadrilateral formed is a rectangle because it has2 pairs of opposite sides equal in measure and the angles are rightangles.

3. Have students find the number of squares within the rectangle ABCD.The number of squares within ABCD is 12. Define area as the numberof square units within a figure.

4. What situations require the computation of area? What situationsrequire the computation of perimeter? List responses and make atable similar to the following:

AREA PERIMETER

carpeting a room installing tackless trips

for carpetingseeding a law

buying fabric for a tablecloth

painting a mirror'ssurface

mirror a wall

fencing around a lawn

baseboard molding around a floor

buying trim for the edge of atable cloth

frame for a mirror

Generalize that area involves covering a surface completely, whileperimeter involves measurement around a figure.

S. Discuss the type of units used to measure area and compare theseto the units used to measure perimeter. Elicit that square unitsare used to measure areas because they are a convenient shape com-pletely covering any surface. Demonstrate this by showing severalfigures as follows and discuss ily each of these would be harder towork with than squares.

Figure 2

1

1°1 .: 3

6. Do Applications A, B, and C.

APPLICATIONS:

A. Determine which of the following are examples of linear measure(perimeter) and which are surface measure (area):

1. the amount of thread on a spool

2. a boy's height

3. a plank, a slat, or a wire fence to be whitewashed

4. a floor to be sanded

S. a fringe to be put on a bedspread

6. a rug to be cleaned

7. a painting on a wall

8. a frame to be put on a painting

B. Find the area and perimeter of each rectangle:

1. . :2.

4.

111

6.

-

...

Figure 3

C. Draw rectangles having the following dimensions. Draw in the squareunits and find the area of each rectangle.

1. 6 by 4

2. 5 by 3

3. 10 by 2

SUMMARY:

Review that a surface or area must be measured with a square unit, whereasa perimeter is measured with a linear unit.

CLIFF-HANGER:

1 1Find the area of a rectangle which measures 32 units by 42 units.

LESSON 2

AIM: To calculate the areas of rectangles and squares


- state that the area of a rectangle can be found using the formulaA = lw (or A = bh).

- state that the area of square can be found using the formula A = s 2.

- use the formulas A = lw and A = s2

(a) to calculate areas and,

(b) to find a missing dimension when the area and onedimension are given.

VOCABULARY:

New terms: length, width, base, height, side

Review: area, formula

CHALLENGE PROBLEM:

Find the area of a rectangle which measures 15 units by 18 units

DEVELOPMENT:

1. Elicit the response to be 270 square units. Discuss the techniquesused by the students to obtain this result. It should be clear tostudents that , while countirg squares will yield the solution,this procedure is not convenient. The same result can be obtainedby recognizing that a rectangle has rows of squares, each contain-ing the same number of squares. The total number of squares canbe found by multiplying the number of squares in each row by thenumber of rows.

18

TOWS

15 squares in each row

113

1 16

Figure 1

IA = lw I

2. If the number of units in each row is called the length (1) or base(b) of the rectangle, and the number of rows is called the width(w) or height (h) of the rectangle, then the area of the rectanglemay be found by multiplying the length by the width dr the base bythe height. The formula which expresses this relationship is:

or I A = bh

3 Use this formula to check the result for the challenge problem asfollows:

A = lw

A = 15 x 18

A = 270

This area is 270 square units.

4. Use the formula A = lw to solve the cliff-hanger problem as follows:

A = lw1 9

A = 37 x -2-

7 9A =T xT63

A .4

3

i

A = 1534

This area is 15 square units.4


6. If the side of a square is 15 cm, what is the area of this square?Elicit from the class that a square is a type of rectangle where thelength and width are the same. If each dimension of the square iscalled a side (s), then the area of the square can be found bymultiplying a side by a side.

The formula which expresses this relationship is: A = s x s or A = s2.(Recall the use of exponents and how to read s2 as s squared or s tothe second power.)

7. Use the formula A = s2to solve the problem in item 6.

A = s2

A= 152

A= 15 x 15

A = 225 cm2


9. Given that the area of a rectangle is 72 sq. inches and the lengthof this rectangle is 6 inches, find the width. Elicit from thestudents that this problem can be solved in the following way:

Given: A = 72, 1 = 6

Using the formula A = lw, substitute72 = 6w

1 144

1. I 7

Solve this equation recalling the division property of equality:

72 6w6 6

12 = w: The width is 12 inches

Check the problem: If 1 = 6 inches and w = 12 inches

A = lw

A = (6) (12)

A = 72 sq in. which was the given area

10. In a similar manner have students find a side of a square whosearea is 49 m2.

A = s2

49 = s2

1/49 = s

7 = s This side is 7 meters

(Recall the meaning of the square root of a number and thesymbol "1/--- ")


APPLICATIONS:

A. Find the area of the rectangles whose dimensions are given:

1. 1 = 8 m, w = 22 m

2. b = 36 mm, h = 17 mm

3. 1 = 3.2 cm, w = 6.3 cm

4. b = 74

in ,h = 22 in.

5. 1 =2 4

yd w = 3== yd

B. Find the area of each square having the given side:

1. 28 m

2. 2.T

ft

3. 1.8 mm

4. 2.25 m

C. Find the missing dimension in each quadrilateral.

1. A rectangle whose area is 325 cm 2and width is 13 cm

2. A rectangle whose area is 3.45 km2

and length is 2.3 km

3. A rectangle whose area is 11 sq. in. and width is 2-:12- in.

4. A square whose area is 400 m .

5. A square whose area is 50 cm 2(answer: SO cm)

6. A square whose area is 16 sq in. (answer: IF= 1 in.)16 4

1151 8

SUMMARY:

Review the formulas for finding the area of a rectangle and a square.

CLIFF-HANGER:

Raphael's back yard is in the shape of a square. He used 196 sq ft of

sod to cover the ground. How many feet of fencing will he need to enclosethis area?

1 ; 8116

LESSON 3

AIM: To solve verbal problems involving area and perimeter


- distinguish between area and perimeter in verbal problems.

- solve verbal problems involving area and perimeter of rectanglesand squares.

VOCABULARY:

Review: area, perimeter, square, rectangle, enclose, cover

CHALLENGE PROBLEM:

Floor tile costs $15 per square meter. What is the cost of covering a

rectangular floor 5 meters (5m) long and 4 meters (4m) wide?

DBELOPMENT:

1. Use challenge problem to elicit answers to the following questions:

a. What information is given? (cost per square meter and dimen-

sions of a rectangular floor)

b. What information is asked for? (Cost of tiling theentire floor.)

c. What information must be found in order to find this total

cost? (number of square meters in the area of this rectangular

floor)

d. What words in the problem indicate the need for finding area?(covering a floor, or cost per square meter)

2. Solve the problem:Area of rectangular floor = 1 x w

Area of rectangular floor = 5 m x 4 m

Area of rectangular floor = 20 m2

20 square meters of tile are needed to cover the floor.

Total cost = Cost per Unit x Number of Units

Total cost = $15 x 20

Total cost = $300

3. Solve the cliff-hanger problem given in the previous lesson byasking questions similar to those in item 1.

Emphasize that the terms "enclose" and "feet" indicate the need

to find perimeter and that "square feet" indicates that the areais given.

117

1 "04/

Given: A = 196 sq ft

Since A = s2

196 = s2

Arii= s

14 = s

Each side of the square is 14 ft. long.

Since P = 4s

P = 4 x 14

P = 56 ft

Raphael needs 56 feet of fencing to enclose his yard.

4. Do applications.

APPLICATIONS:

Before obtaining the solutions for the following problems, the teacher

should have students underline in each problem the key words or phraseswhich indicate the need for finding area or perimeter.

1. Nancy is making a holiday tablecloth and wishes to trim it withfringe. The dimensions of the finished tablecloth are 111 yardsby 2 yards. The cost of the fabric is $3.99 a square yard, andthe cost of the fringe is $2.49 a yard.

a. What is the cost of the fabric?

b. What is the cost of the fringe?

c. What is the total cost?

2. Gerald intends to wallpaper one wall in his hallway. The wall is

8 ft high and 15 ft long. Each roll of wallpaper will cover30 sq ft How many rolls will he need to cover the wall?If each roll costs $12.95, what is the cost of wallpapering thiswall?

3. If grass seed costs $5 a pound, and one pound of grass seeds covers2500 square feet, how much will it cost to seed a rectangular lawn20 ft by 50 ft ?

SUMMARY:

Review key phrases or terms which indicate whether area or perimeter is

required.

118 1 21

CLIFF-HANGER:

An L-shaped room has the dimensions shown:

24 ft

V------18 ft

30ft

15 ft

If carpet costs $12.99 a square yard, how much will it cost to carpetthis L-shaped room?

119 122

LESSON 4

AIM: To solve problems involving areas of complex figures


- calculate areas of complex figures involving rectangles by dividingthe figures into component rectangles and adding.

- solve problems involVing figures of this type.

VOCABULARY:

New terms: complex

Review: dimensions, rectangle

CHALLENGE PROBLEM:

Find the area of Figure 1,

Figure 1

24'

DEVELOPMENT:

18'

30'

15'

1. Elicit that the figure in the challenge problem can be dividedinto 2 rectangles in either of the following ways:

Figure 2 Figure 3

12'14(--- 18

12'118'

-r-

9' A9'

24' 24' C

15' 15' D

12' 18'

1*- 30' 4.1

120

if2 3

15'

2. Determine the dimensions of each of the smaller rectangles.

Tn figure 2 rectangle B is 30' x 15' and rectangle A is 9' x 12'.Have students explain how the dimensions of rectangle A areobtained.

Have students calculate the areas of rectangles A and B.

Area of A = 9 ft x 12 ft Area of B = 30 ft x 15 ft

Area of A = 108 sq ft Area of B = 450 sq ft

Elicit that tL total area is the sum of A and B or 558 sq ft

3. Similarly, calculate the areas of C and D in Figure 3 to be 288 sq

ft and 270 sq ft respectively, whose sum is also 558 sq ft

4 Have students state that in order to find the area of a complexfigure made up of rectangles, it is necessary to divide thefigure into its component rectangles. The dimensions of eachrectangle must be found and the areas calculated. The total areaof the figure is the sum of the areas of the component rectangles.

S Do Application A.

6. Return to the cliff-hanger problem.

Elicit that to find the cost of carpeting the L-shaped room, thearea of the room must first be found by using the method describedin item 2 or 3. (Since the price is given for square yards, thedimensions of the room should be changed from feet to yards beforeany calculations are done). Once the area of the room has beencalculated, the number of square yards is multiplied by the costper yard.

-I- 4 yd

3 yd A

8 yd

Figure 46 yd .....,1

6 yd

5yd B

F3 yd

10 yd

Total area = 12 sq yd + 50 sq yd

= 62 sq yd

Cost of carpeting room = 62 x $12.99.

Cost of carpeting room = $805.38.

121

124

3.


APPLICATIONS:

A. Find the areas of each of the given complex figures whose dimensions

are shown:

1.

3.8 ca,

Figure 5

1.5 et-et

2.9 cry

4.

2. 14- 9.5 ret ---zi

i

11

B. Figure 6 shows a wall with a door and a window. What is the area of

the window? What is the area of the door? If the wall is to be

painted, how many square feet of surface need to be covered? (The door

and window won't be painted.) If the 2 side walls of the room are 10

ft long and there are no other windows or doors in the room, whatis the total surface that needs to be covered with paint?

8 ft

Figure 6

3 ft 5 t

7 ft

3 ft

12 ft

(Note to Teacher; Find other irregular figures of this type and

compute areas. Use the Real Estate section of a Sunday paper.)

SUMMARY:

Review the methods used to compute the areas of complex rectangular

shapes.

122 .

2 51

CLIFF-HANGER:

As shown in the diagram, a flag is to be made in two colors, red and green.

If the dimensions of the flag are 30 in. in length and 20 in. in width, what

is the area of the red cloth needed?

I

RED

GREEN

30 in.

12326

20 it

LESSON 5

AIM: To find the area of a triangle


apply the formula A = 1bh to calculate the area of right triangles andexpress the area in square units.

- identify the base and height of any triangle where the bottom sideis the base.

- use the formula A = 1/2bh to calculate the area of any triangle.

use the formula A = !ibh to calculate the base or height of a triangleif the other dimensions are given.

VOCABULARY:

Review triangle, right triangle, diagonal, rectatigle, base, height

CHALLENGE PROBLEM:

Find the area of the figure shown.

DEVELOPMENT:

1. a. Have students suggest ways of finding the area in the challengeproblem. (counting)

h Students should realize from their work with rectangles thatcounting is not always practical.

c Elicit that if this right triangle were seen as half of arectangle, we could multiply 1- x 10 x 6 and the area wouldbe 30 square units.

d Have pupils build up a rectangle on the original right triangleas shown. Figure 2

i

124 127

(Note: This construction should be demonstrated at the boardor on the overhead projector.)

e. Elicit that the area of the right triangle is equal to 1/2 the

area of the rectangle. (Pupils may cut along the diagonal ofa rectangle and place the triangles on top of each other showingthat the figures are equal in area.)

f. Elicit that since a formula for the area of a rectangle is

A = bh and the area of a right triangle is 1/2 the area of arectangle, a formula for the area of a right triangle is:

A = 1/2bh

2. Use the formula for the area of a right triangle to answer thecliff-hanger problem and arrive at the result 300 square inches ofred cloth.


4. a. In Figure 3,/iABC has a base of 75 feet and a height of 50

feet. Elicit the meaning of height (as in tree, building,etc.) as the perpendicular distance from the highest point to

the bottom (base).

Figure 3

B

SinceAABC is not a right triangle, ask the pupils if the for-mula A = 1/2bh can still be used to find the area ofAABC? Why?

b. Develop the concept that the formula A = 1,h may be used tofind the area of any triangle by constructing a rectanglearound triangle ABC as in Figure 4 and proceeding as follows:

Figure 4

125 i 28

1.

(1) The area of AI = the area of QII and the area ofAIII = the area of AIV. (Refer to Figure 2 inorder to explain this.)

(2) Since-AI = AI1 and

Athen /AI + III = All + AIV using the additionproperty of equality.

(3) Since AI 4 AI II = ZNI1 AIV and all four trianglesadd up to the entire rectangle, each sum (LEIand (6II +LIV) is one half of the rectangle.

(4) Therefore the area of 4.6I +Lail = 1/2 the area of

the rectangle

or

the area of ABC = 1/2bh

S. Use the formula A = 1/2bh to compute the area of d ABC.

A = 1/2bh

6.

75 SOA = 1/2 x x

1 1

A = 1875 sq ft

Do Applications A, B, C, and D.

APPLICATIONS:

A. Find the area of each of the right triangles pictured:

5.5 cm

1. B

31/2 21'4

B. Name the base and the height of each of the triangles drawn:

D C

2. H

E F G T

126

129

K

C. Find the area of each of the triangles in Application B (Figure 6)if the lengths of the line segments are given.

1. BD = 12.6 m, AC = 9.5 m

2. FG = 5 in,) HE = 41/2 in.

(Point out that for this type of triangle, the lengthof the base does not include its extension outsidethe triangle.)

3. TL = 18 ft, MK = 64 ft

D. Complete the chart below by finding the missing dimension in each

triangle.

a.

b.

c.

d.

SUMMARY:

Base Height Area

28 cm ? 140 cm2

? 60 m 720 m2

21 in.? 42 sq in.

? 10 cm 2.5 cm2

Review the meaning of the terms base and height of a triangle and theformula for finding the area of a triangle.

CLIFF-HANGER:

One room at the corner of a large building has a floor shaped as shown.If tile costs $3.00 a square foot, find the cost of tiling this room.

TI

1

I

I

12 f'.

15 ft

127

LESSON 6

AIM: To find the areas of complex regions involving rectangles and triangles


- divide complex regions given in a diagram into rectangles and triangles

- calculate the areas of complex regions involving rectangles andtriangles.

CHALLENGE PROBLEM:

One room at the corner of a large building has a floor shaped as shown.Find the area of this floor.

4 yd

DEVELOPMENT:

1. Using the challenge problem, elicit that the area of the floorcan be found by finding the sum of the area of the rectangle andthe area of the triangle. The students should obtain a result of20 square yards for the area of the rectangle plus 8 square yardsfor the area of the triangle. Therefore the area of the floor is28 square yards.

2. Solve the cliff-hanger problem by eliciting that the area of thefloor can be found in the same way as was shown in the challengeproblem. The students should obtain a result of 180 sq ftfor the area of the rectangle and 72 sq ft for the area ofthe triangle for a total of 250 sq ft.

Therefore: The cost of tiling the floor = 250 x $3.00.

Cost of tiling the floor = $750

3. Do applications.

APPLICATIONS:

A. Using the given information in each of the following diagrams, findthe total area of each figure:

128

1.Figure 2

<-- / 5" ----->

IS"

i

4,

2.

3. 111+"

>144--T\ A

II

s. 14_

-f r5. 5 oi,

4, /---tI .5 on

J., L

14--3.1-4c,-A4,1

I

1E-- ------12.8 o-n- -

129 i .(2

B. A barn has a shape as in Figure 3. Using the information given inthe diagram, find the cost of painting the front, back and walls ofthe barn if a gallon of paint costs $15.95 and covers an area of20 sq ft.

(Note to Teacher: Discuss the need for purchasing a full gallon eventhough only a part of it will be used.)

SUMMARY:

Review the method of finding the area of complex regions involvingrectangles and triangles.

CLIFF-HANGER:

A revolving sprinkler sprays a lawn for a distance of 7 m. How manysquare meters does the sprinkler water in 1 revolution?

130

)

LESSON 7

AIM: To find the area of a circle


state the formula used to find the area of a circle: A = Irr2

.

22- apply the formula to find the areas of circles using Tr= --

VOCABULARY:

Review: circle, radius, pi, diameter, circumference

CHALLENGE PROBLEM:221

Find the circumference of a circle whose radius is 31/2 m fuse = --j7

DEVELOPMENT:

1. Use challenge problem to review diameter, radius, and circumference

of a circle. Recall meaning of circumference and the formulas used

to find circumference of a circle:

C = 2Trr

C = ir d

Review that it is the ratio of the circumference to the diameter

in any circle and approximately equals or 3.14.

2. Solve the challenge problem as follows:

C = 2 Trr

22 1

C = 2x .-7 x 3.f

12, 22 .?".

C= y x --;---r

1

x..?

1

C= 22m

3. Recall the cliff-hanger problem. Elicit that the students must findthe area of the circle in order to solve the problem; however,they have not yet learned a method for doing so.

4. Develop the formula for finding the area of a circle in the

following way:

a. Draw a circle and divide it into 16 equal parts as in Figure 1.

b. Demonstrate that the circle may be cut apart and put back

together as in Figure 2. This can be done effectively on the

overhead projector.

c. Figure 2 is similar in appearance to a rectangle. However a

closer approximation can be obtained by cutting the sector on

the far right in half and placing this half on the left side

as in Figure 3.

131

Figure 1

Figure 3

Figure 2

1,-------2I C., IT X r___ .._ _ .1

I

d. Elicit that the dimensions of the rectangle formed in Figure 3are approximately 1/2C (the base) and r (the height); therefore,the area of the rectangle is approximately given by:

A = bh

1A = -2; C x r

e. Recall that C = 2 ir r, so by substitution1

A = T (2 rr r) x

This

or1 ?"' .1,..A = u xrxr

can be written in simplest form as:

A = D'" r2

This is the formula for the area of a circle in terms of itsradius.

5. Solve the cliff-hanger problem as follows:

A = ir r2 where r is 7 meters

22A . -7 x 7

2

6. Do applications.

A22 .7-1 7t k = -5,.- X -,- X T

l

A= 154 m2

132

APPLICATIONS:

A. Find the areas of circles having the following radii. (Use Tr= 22)

1. 14m

2. 20cm

3. 42yd

6

4. 1/2 foot

S. 3-,T

B. Find the areas of circles having the following diameters (Note: Reviewfinding the radius when a diameter is given.)

1. 84m Use Tr.= 3172. 7mm

3. 211 ft

C. Solve each of the following problems (use Tr=22

)

1. If an FM radio station can broadcast within a radius of 91 km,over how large an area can the signal be received?

2. From his tower a forest ranger can see a distance of 28 km in any

direction. How many square kilometers can he patrol from his tower?

SUMMARY:

Review the meaning of pi and how to find the area of a circle.

4I.ab

LESSON 8

AIM: To apply the formula for the area of circle


- apply the formula A = 1Yr2 to find the areas of circles, using

11-= 3.14.

CHALLENGE PROBLEM:

A circular swimming pool is 18 feet in diameter. A plastic cover is tobe purchased to fit this pool. If the cost of plastic is $.35 per squarefoot, what will be the cost of the cover to tne nearest dollar?

DEVELOPMENT:

1. Discuss the nallenge problem. Elicit that the area of the covermust be tound in order to find the cost. Have the studentsrecognize that _2_2 is an inconvenient approximation forte in this

Elicit7that another apnroximation for Tr is 3.14, byhaving students convert the fraction 11 to a decimal.

72. a. Using the value Tr = 3.14 find the area of the pool cover.

Since the diameter = 18 feet, r = 9 feet.

A = it r2

A= 3.14 x 9 x 9

A = 254.34 square feet

b. The cost of the cover cost per sq. ft. x number of sq. ft.

Cost = $.35 x 254.34 sq.ft.

Cost $89.019

The ost to the nearest dollar is $89.

3. Do applications.

APPLICATIONS:

A. Find the area of the circle described using Tr 3.14

1. r = 6 cm

2. r= 8 cm

3. r= 2.2 m

4. r = 0.7 mm

B. 1. Would you get more pizza if you bought two 10-inch pizzas or one16-inch pizza?

2 Which is larger: the area of a circle 6 cm in diameter or the areaof a square whose side is 6m? How much larger?

3. A circular walk surrounds a fountain is the park as shown:

134

t)

Figure 1

F represents the fountain. The

walk, which is 6 ft wide, is repre-sented by the shaded region. If

the cc't of concrete is $2.34 asquare foot, what is the approximatecost of constructing this walk?

4. Find the total area of each of the following figures:

Figure 2

5

20"

20"

5. A football field consists of a rectangle with semicircles at

each end, as shown. Using the dimensions given, find, to thenearest square yard, the number of square yards of sod needed to

cover the field. (Use 7r= 3.14.)

Figure 3

SUMMARY:

100 yd .

I

40 yc

Ex. 13

Review the meaning of area, the formulas used for finding the areas ofrectangles, squares, triangles, circles, and the method of finding the

area of a composite figure. Compare the units used to measure area with

those used to asure perimeter.

135

UNIT XI. Indirect measure and ScalingLESSON 1

AIM: To distinguish between direct and indirect measurement


- describe situations in which a measurement of length or distance

cannot be made directly.

- apply formulas to find measurements of quantities indirectly by using

measurements that have been made directly.

VOCABULARY:

New terms: direct measurement, indirect measurement

Review: formula

CHALLENGE PROBLEM:

How far away is a f:ash of lightning which is seen 5 seconds before the

thunder is heard?

DEVELOPMENT:

1. Elicit from class that the challenge problem requires finding adistance or measure of length. Suggest the possibility of obtain-ing the rez,,,it directly by measuring the distance from the pointwhere the thunder is heard to the point where the lightningactually struck.

2. Discuss the meaning of direct measurement (where the unit of mea-sure is applied rectly to the quantity being measured). List

objects whose measure can be found directly suc.. as dimensions of

a room, height of a person, length of a piece of fabric or length

of a piece of wood.

3. Elicit that to measure the distance lescribed in the challengeproblem directly is possible hilt very difficult. Another method

of solving this problem is indirect measurement. One type of

indirect measurement is through the use of a formula. One formula forfinding distance is d = rt. Since the rate or sreed of sound(in air) is approximate'y 1100 feet per second and it took 5seconds until the thunder was heard:

d = rt Note: Since the speed of light is

d = 1100 x 5186,000 mi/sec, it is disregardedin this problem.

d = 5500 ft

Since 5280 feet = 1 mile, the lightning was a little more than one

mile away.

4. Elicit situations in which direct measurement would be impossibleor impractical; for example: the height of a mountain, or build-

ing, or a flagpole; the distance across a body of water, orbetween the earth and the other planets, or between cities.

137

S. Discuss the fact that most measurement is indirect. For example,when you weigh an object, you are usually finding the stretch ofa spring; when temperature is measured, you are measuring theamount of expansion or contraction of metals; when time is mea-sured, the amount of movement of gears is being measured. One ofthe few measurements that can be made directly is that of length.

6. Discuss situations in which there is a choice as to the type ofmeasurement used. For example, the area of a rectangle can befound directly by counting the number of square units it containsor indirectly by multiplying its length by its width. Elicitthat indirect measurement is often more convenient - easier, faster,and more efficient.

7. Do applications.

APPLICATIONS:

1. Determine which of the following measurements can be done directlyand which must be done indirectly:

a. finding the height of Mount Everest

b. finding the length of a corridor in school

c. findi3ig the diameter of a dime

d. finding the height of the teacher's desk

3. finding tie depth of the Pacific Ocean at the deepest part

f. finding the distance from the earth to the sun

g. finding the heigLt of the school flagpole

h. finding the temperature of a blast furnace

f'naing the weight of al. elephant

some cases, there may a choice between direct and indirect,urement. If so, ask the pupils which would appear to be

preferable.)

2. What direct measurements do we need to know to find each of thefollowing indirectly?

a. the area of the floor of a classroom

b. the -Primeter of a triangle

c. the ircumference of a circle

d. the area of a triangle

3. Radar trax_Ils at 186,000 miles per second. If it takes 2.8 secondsfor a radar signal to bounce ciff the moon and back to earth, whatis the distance from the earth to the moon? (Use d = rt, Note ittakes 1.4 seconds for the signal to go fccm the earth to the moon.

4. A merchant wishes to set up ; display of canned goods. How manycans will he need for the display if he wishes to place 9 cansin t'-,e bottom row and 1 can less in each row until there is only

1381

1 0,

1 can in the top row? Use

the formula S = 1.4 (a + I) where

S = the sum, n = the number of

rows, a = the number in the bottom

row, and 1 = the number in the

top TCW.

Check the result of the indicrect measurement by directly countingthe number of cans.

5. Henry borrowed a book from a library. The library charges 15¢ forthe first 3 days and 2¢ a day for each additional day. If Henrykept the book for 19 days, how much did he have to pay? (Usethe formula: c = 15 + 2(n-3) where c = the charge in cents andn = the number of days he kept the book.)

6. How high is a building if a rubber ball which is dropped from theroof takes 2 seconds to hit the ground? (Use the formula s = 16t2where s = the distance (height) in feet and t = the time inseconds.)

SUMMARY:

Review the difference between direct and indirect measurement and theneed for the latter in some measurement problems. Pupils should under-stand that indirect measurement requires the measurement of some quantityor quantities directly and the performing of calculations using thzsedirect measurements to obtain the desired results.

CLIFF-HANGER: See next page.

139 141

CLIFF-HANGER:

A treasure map shows that a treasure is buried 60 paces due east ofthe BIG ROCK. If Stan is 6'4" tall and Oliver is 5'6" tall and eachlooks for the treasure, will they dig in the same place? Who will findthe treasure?

GO paces

140 142

-----...

B GRoc-

I

'SON 2

AIM: To understand the meaning am. use of scale drawing

PERFORMANCE OBJECTIVES: Students kill be able to:

- state a need for standard units of measure.

- use a scale to determine the actual number of unit, represented in adrawing.

- measure the length of line segments using a ruler

determine an actual distance given a unit scale and a scaled length.

VOCABULARY:

New terms: scale, scale drawing

Review: unit, indirect measurement, direct measurement

CHALLENGE PROBLEM:

Tom and Jerry found the following treasure map:

`.

How far r.7om the rock must Tom and Jerry dig?

DEVELOPMENT:

NA

= 1 5 c.e..et.

E

1. Discuss the challenge problem. Compare it with cliff-hanger problem.How do the two maps differ? Elicit: the scale on the challengeproblem gives a standard unit of measure whereas the map in the

141

cliff-hanger problem uses paces (a non-standard unit of measure-ment which differs from person to person).

2. Relate the scale 15 ft) to the scale on a pictograph.How can the distance from Big Rock to the treasure be found?Elicit the need for finling the number of units in the length ofthe line joining the Big Rock to the treasure on the map.

B1 G

Roc k 11-------4 t 4------ ---- 4-- -PI

A

> C

7: 15 -cc &.

Have students determine that the line contains JO of the givenunit lengths. Therefore, since each unit length represents 15feet, the line is 15 (10) or 150 feet long.

3. A map of this type is called a scale drawing. Have students lie.other examples which involve scale drawings or scaling. Elicitsuch examples of scale drawings as: road naps, blueprints, patternsin dress-making, model making (airplanes, cars), enlargements (blow-ups). Elicit that actual measures can be founo indirectly by mea-suring a scaled figure directly and knowing the value of the scale.

4. Do ar'lications.

APPLICATIONS:

A. Measure each line segment and write the distance using the scale ofmiles or kilometers given. (See Figuro 3)

142 I 4 ,1

APPLICATIWS:

A. Measure each line segment and write the distance using the scale ofmiles or kilometers given.

Figure 3

1. 1 inch = 200 miles 2. 1 inch = 80 kilometers

F-

4

1

3. 1 inch = 50 miles I 4. 1 inch = 20 kilometers

F--I

F-

143 :45

Figure 4

Distances Between Cities___,

MI LIU f. 4? 01_1'

DEt,IVIZST. WU'S

NEwORLEAWS

cr)tv

N1Qw Yoat;E

oin0 (NI*

SCALA: Jur. /000 MIL6S

MI AM1

00I()I

000-A

B. Using the map in Figure 4 record the distan e between cities in the chartbelow. (Measure to the nearest 4 in.)

DENVER

LOS ANGELES

MIAMI

MILINEAPOLIS

N!EW ORLEANS

NEW YORK.

SEATTLE

ST. LOUIS

AP,144

,./

Figure 5Major U.S. Cities

\A./ASHIUGTOkl

D.C.

147

SCALE IN KILOMETRES

O 26o 11,60 660 800

1 cm :--- 2.00 Km

LIi 4 8

C. Figure 5. (Note to Teacher: Students may use metric ruler below map.)

1. Using the given scale, find the approximate distance between eachpair of cities listed.

a. Salt Lake City to Denver

b. Tucson to Tulsa

c. Chicago to Cleveland

d. St. Paul to Houston

2. Which of the following trips is longer? How much longer?

Los Angeles -) Chicago ------4 Washington

Los Angeles ------4 Atlanta --) Washington

SUMMARY:

Review how to use a scale and a scale drawing to determine actualmeasurements.

CLIFF-HANGER:

Wendy saw the following design pictured in a book. She decided to usethis design for a belt that would be 1 inch wide. How would she accom-plish this?

146 1 4 9

LESSON 3

AIM: To identify similar figures


- state tnat the enlargement or the reduction ratio is called a scale.

- compare any two quantities with a ratio.

- state that enlargements or reductions create figures similar to the

- state that when figures are similar, the corresponding parts (lines)are in proportion.

identify a true proportion by:

1. comparing fractions.

2. using the proportion propeity.

VOCABULARY:

New terms: proportion, means, extremes, similar, terms, corresponding

Review: scale, ratio

CHALLENGE PROBLEM:

Which of the following pictures represents an enlargement of Figure A?

A. [5-

1.

2.

3.

147i

DEVELOPMENT:

1. Discuss the challenge problem. Elicit from students that in Picture 1the length was increased (doubled) but the width was kept the same.This resulted in a distortion of the original diagram. Similarly,

in Pf..Lture 2,the width was increased (doubled) but the length

was left the same. This re.:,t1te in another distortion o: thesprig! al diagram. However, in I-igure 3 both the length and thewidth were increased (doubled) maintaining the same shape as theoriginal design.

Explain that figures having the same shape are called similarfigures. Therefore, Figure 3 is similar to Figure A.

2. Have students measure the length and width of Figure A and the..ength and width of Picture 3. Have students write the ratio ofthe width of Firure 3 to the width of Figure A (2 to 1 or 2:1 or y).Ask students to hypothesize the ratio of the length of Figure 3to the length of Figure A. Verify the hypotheses by measuring. Explainthat the ratio of corresponding lines is called the -nlargement scale.

3. When figures are simil the enlargement or reduction scale is thesame for all corresn-:, ...mg lines. Verify this fact by measuringa pair of co-responding lines (other than length or width) frome-e patte-rns, and writing the ratio of the enlargement part to theoriginal part. Elicit that in this challenge problem each r7...tio

could be simplified to 2:1

4. Define proportion as a statement of 2 equal ratios. Have stye,.

write proportions such as:

Length of Figure 3 Width of Figure 3Length of Figure A Width of Figure A

or

Width of Figure 3 Any part of Figure 3Width of Figure A The corresponding part in Figure A

Therefore, when two figures are similar, their corresponding partsare in Proportion.

5. Do Application A.15

6. Do the ratios TT and To- form a proportion? Have students recall

Lne definition of a proportion (a statement that two ratios are

equal) and u_a this definition to write the following: Does9 15

?IT=

20

Lead students to answer this question by the two methods described

below:9

a. DoesIT = 15

? Method A: Rewrite each ratio

15

12

Dividing numerator anddenominator by 3 resultsin the ratio

4

148

lowest terms:

Dividing numerator anddenominator by 5 resultsin a ratio _1

4

111

r------

Therefore 19 is a true proportion.ISii

2 20

b. Does =15

192 20

Method B involves first defining the terms of a proportion.The four numbers that form a proportion are called the terms.In this proportion 9 is the first term, 12 is the second term,IS is the third term, and 20 is the fourth term. In anyproportion the second and third terms are called the means andthe first and fourth terms are called '..he extremes. Thefollowing property is used:

In a true proportion)tin product of the means is equalto the product of the extremes.

Illustrate the proportion property as follows:

9 15i

12Ti f 12 x IS = 9 x 20.

15Since 12 x 15 = 180 and 9 x 20 = 180, the proportion I2

z 20is a true proportion.


APPLICATIONS:

A. 1. Name the pairs of figures which appear to be similar.

Figure 2.

2. Tell which pairs of figures appear to be similar in figure 3.

a.

g.

Figure 3

b. c.

e. f.

j.k.

B. Which of the following are true proportions?

, 3 9 19 57" T 12=3

3 9

2.2

=5 5

7 16

11

9 20

L n

5.10---15

2

7

.

=

4

6

10

35

C. For each pair of similar figures, write a proportion which relatestwo pair of corresponding sides. Show that each proportion is atrue proportion.

150

1 53

1.

3.

8

A

F

A!,'

5. V0,...\ N.--------____ a 6.

i \ n\ \ 1

I 12 \ 81

\ I 3 I \\

AlI \\

N

z 20

24

X

Figure 4

2.

G

4.

\12

15

4.5 16

SUMMARY:

0

5Al

//15/ I

/ 100

8

121/2\\\!X

Review the meaning of ratio and scale. Review the definitions of similarfigures, proportions, and the terms of a proportion. Have studentsstate the two methods used to ',how whether a proportion is true.

CLIFF-HANGER:

The shadow of a tree is 16 feet long at the same time that the shadowof a boy is 4 feet long. If the boy is 5 fec tall, how tall is thetree?

151 1 54

LESSON 4

AIM: To use proportions to find unknown sides of similar triangles


- use lte proportion property to solve proportions (e.g.,b! =

c).

- write and solve proportions to find an unknown side in similar triangles.

VOCAPLARY:

New terms: perpendicular

Review: similar, proportion, means, extremes, division property ofequality, indirect measurement

CriALLENGE PROBLEM:

Find the value of x that makes the proportion true.

x 16

5 4

DEVELOPM, 'TT:

1. -tudents may obtain a solution to the challenge problIm in severalways. Elicit the following :

a. Recall the cross-product property of true proportions. Name

the means (5 and 16) and the extremes (x and 4) and write the

products 5.16 and 4x and the equation relating these products:

80 = 4x

b. Recall the use of the division property of equality to solvethis equation:

80 4x4 4

20 = x

c. Check that 20 is the solution.

By reds By proportion

20 16 20 16 9

5 4 5 4

4 = 4 5.16 = 4.20

2. Do Application A. 80 = 80

3. Discuss the cliff-hanger problem.

a. Guide the pupils to realize that on a sunny day a tree, aswell as a building, people, and all kinds of objects ir. thestreet, cast shadows. Since these shadows are on the ground,it is possible to measure them quite easily with the usualinstruments for measuring length (rulers, tapes, etc.). Thelengths of shadows change as the sun rises and sinks in the

152 155

sky, so comparative measures of shadows must always be madeat the same time of day.

b Have students draw a picture representing the situation describedin the cliff-hanger as in Figure I. The dotted lines indicatethe path of the sun's rays. Since all measurements are beingmade at the same time of day, the rays strike the ground atthe same angle in both objects pictured.

Figure 1

/6P1S s

// /,4

t.(,

,'V;

///

,,Lfri -/s

3/ di93

.)....,f

16

NOTE: Diagramsare not drawn toscale.

c. Elicit that the physical situation can be represented by ageometric model as In Figure 2. The students will note thatboth triangles are right triangles since the tree and thepupil standing next to the tree make right anf,les with (areperpendicular to) the ground. Because of the nature of theway the sun's rays strike, students should note that thetriangles formed are similar.

4

Figure 2

16

x

NOTE: Diagramsare not drawn toscale.

d. Nave students recognize that the cliff-hanger problem actually:equires them to find the .issing leg in the larger of the twosimilar triangles. Sinct the triangles are similar, thecorresponding sides are in proportion.

e. If the unknown leg of the triangle is represented by x, havestudents write a proportion relating the corresponding sidesof the triangles.

x 16

T 4

1.

f. Students sho'ild recognize this proportion Es the same onesolved in the challenge problem. The solution to this equa-

tion is = 20. Therefore, the height of the tree is 20 feet.Elicit that the height of the tree was formed indirectly bysolving a proportion resulting from similar triangles.


APPLICATIONS:

A. Find the missing term in each of the following proportions.

8 x , 1

1'= --

16 64. .,

14 42i 6

2. =x 21 x 2.4

=3.x 9_ _

6 g- =

36 4 3 6

=9 x_ 7' 1.4 Z4*

36 4 '=

x8

3.6 1.5

B. Each of the following pairs of figures are similar. Find the indicated

side.

r 9km

/ in

7 cm

15 1,:n

x

4 niL J

/3 5 cro 4 cm

Figure 3

2.

125 in

6 ydr 9 yo

(C. 8 ft

154

4 It

9 cm

12crn

x

11 ft

5 7

C. 1. A tree casts a shadow 120 feet long at tht same time as a manstanding next to the tree casts o shadow 8 feet long. If theman is 6 feet tall, what is the height of the tree?

2. An apartment building casts a shadow 75 feet long at the sametime as a telephone pole next to it casts a shadow 15 feet long.If the pole is 20 feet high, what is the height of the building?

SUMMARY:

Review the principles upon which indirect measurement using shadow isbased: the triangles created by the shadows are similar to each otherand the corresponding sids of similar triangles are in proportion.

CLIFF-HANGER:

On a typing test Alicia typed 80 words in 4 minutes. At the same rate,how many words would she type in 10 minutes?

155 1

LESSON S

AIM: To use proportions to solve problems involving ratios and rates


- represent relationships in verbal problems as proportions involvingrates or ratios.

solve proportions using the proportion property.

VOCABULARY:

Review: rate, ratio, proportion

CHALLENGE PROBLEM:

In a school the ratio of the number of girls to the number of boys is2 to 3. If 630 boys attend the school, what is the number of girlsattending the school?

DEVELOPMENT:

1. Elicit that many situations besides similar figures lead toproportions. Since this problem involves ratio, its solu-tion can also be found by solving a proportion. Translatethe given information as follows:

number of girls 2

nmoer of boys 3

Using the additional information given in the problem, obtain thefollowing :

number of girls 2

633

Use a variable to represent thc, number of girls to simplify theexpression:

g 2

630 3

Solve the proportion using the proportion property.

3g = 2 x 630

3g 1260

3 3

g = 420

Therefore, the number of girls attenuing this school is 420

2. Discuss the cliff-hanger problem. What ratio is implied by thestatemcnt "Alicia typed 80 words in 4 minutes"? (words per minute

wordsorminutcs

). Guide students to write the following proportion:

number of words _ 80

;:umber of minutes 4

156 i j

Using the additional information from the problem and "w" torepresent the number of words, obtain the proportion:

80

10 4

obtain the result w = 200. Therefore, Alicia can type 200 wordsin 10 minutes.

3. Pose the problem: If 10 oranges cost $1.29, what is the cost of8 oranges at the same rate? Elicit the following:

number of oranges 10

cost 1.29

Using the additional information and "citto represent the cost,obtain the proportion:

8 10

c 1.29

Obtain the result c = 1.032 round to $1.04. Therefore, the

cost of 8 oranges is $1.04.

4. Do applications.

APPLICATION:

(Note to the teacher: Have students solve each problem using thesame procedure outlined in the development.)

1. A car travels 165 miles in 3 hours. At the same rate, how longdoes the car take to travel 275 miles?

2. Two candidates in a school election split the vote in a 3:2 ratio.The loser received 182 votes. How many votes did the winnerreceive? (Note: 3:2 is the same as

2.)

3, A picture 11 in. x 14 in. is to be reduced to a wallet sizepicture whose longer side is 31/2 inches. Find the shorter sideof the wallet size picture.

4. A recipe to make 4 dozen cookies requires 1/2 cup of butter. Howmuch butter is needed to make 10 dozen cookies?

5. If three cans of soup cost 35 cents, what is the cost of twodozen c-ns at the same rate?

6. In East High School, the ratio of girls to boys in a class is4:5. If there are 20 boys in this class, how many students arein this class? (Hint: How many girls are in the class?)

SUMMARY:

Review the steps needed to solv_ a rate or ratio problem using propor-tions.

CLIFF-HANGER:

On an airline map, the scale is 4 "= 100 miles. If the distance fromLos Angeles to Denver measures 21", what is the distance in miles letweenthese cities?

157 160

LESSON

AIM: To use proportions to solve problems involving scaling


- represent the relationships in problems by using proportions.

- solve problems involving scaling by using proportions.

VOCABULARY:

Review: scale, scale drawing

CHALLENGE PROBLEM:1

In drai:ing plans an architect uses the scale g = 1 foot. How long a linemust he draw to represent a 40 foot wall?

DEVELOPMENT:

1. Use the challenge proble.d to elicit that a scale drawing is similarin shape to the object it represents. The scale or legend estab-lishes the ratio between the scale length and the actual length.Therefore, the following proportion may be used to solve thechallenge problem:

1

8 scale length1 foot actual length

2. Using the remaining information, the challenge problem can besolved as follows: (use s to represent scale length)

1

8 s

1 40

1 40is T x

s =

Therefore a S in. line must be drawn to represent a 40 ft wall.

3. Recall the cliff-hanger and solve it in the same manner as the

challenge problem:

1 .

T in.scale length

100 mi actual distance

Using d to represent actual distance:

1. 2T1

100 d

2 d1 =8

x 1004

158

4 1, 17 1.131125,3. Txriu = xr.

124. d = 850

Therefore the distance between Los Angeles and Denv'r is 850 miles.


5. Pose the problem: Jeff has to make a scale drawing on standardconstruction paper, 81/2" x 11", of his 30-foot long by 25-footwide classroom. How can he obtain an appropriate scale for thisproject?

a. Recall that a scale is the ratio of the scale length to theactual length.

b. Write a ratio so that the length of the scale drawing will just

fit the length of the paper11 in.

. Elicit from pupils that30 ft

such a scale would be impractical because it does not allow for11

drawing the lines and also because To- is an inconvenient fraction.

Elicit that10 in.

30 ftwould be a better choice (this is not the

1 in.only choice) because the scale could be reduced to or

3 ft1 in. = 3 ft.

c, Using this scale determine the scale width of the room. Willthe entire drawing fit on the paper?

1=

3 25

3x = 25

x =8336. Do Application C.

APPL).CATIONS:

1The scale width is 8.T in. which is less

ithan the given 8-2- in., therefore the

picture will fit.

A. 1. If a distance of 80 miles is represented on a map by 21/2 inches,how many miles are represented by 2 Liches?

If on a blueprint of a house a $ inches long represents 6 feet,

what is the lerrth of a line which represents a distance of 32 feet?

3. If a scale is 4 in. = 2 ft find the area of the flGor of a roomwhose dimensions are 21/2 in. by 21/4 in.

4. In a diagram of a computer component, a square silicon chip is8 cm wide. If the diagram is 10 times actual size, how wide is thechip?

159 1132

IiB. In the given floor plan the scale is -a -n. = 1 ft. (See figure 1)

Figure 1

Bedroom 2C.,6+

In0U

Bath Hall

Bedroom 1

U

Farnily room Kitchen

T I

jClosetT (

f

Living room

Dini,:j room

1. Find the actual dimensions of the dining room.

2. What is the number of feet across the front of the house?

3. How many square feet of closet space does the house contain?

4. What is the cost of tiling the hall with 1-foot square tileswhich ...ost $3.50 per tile?

C. 1. If 80 miles is represented by 21/2 in., how many miles are representedby 1 in.?

2. Find the scale if the actual length is 350 miles and the scale lengthis 7 in.

SUMMARY:

How can proportions be used to measure indirectly?

.0 11)3

UNIT XII. Solid GeometryLESSON I

Note to Teacher: All lessons in this unit will be more meaningful if modelsof the solids as well as common objects having these forms are brought toclass.

AIM: To introduce solid geometric figures and the concept of volume

PERFORMANCE OBJECTIVES: Students will he able to:

- identify plane and solid objects as rectangular, square, circular,and triangular.

- distinguish between drawings or objects which are plane or solidfigures.

- state that a prism i3 named by the shape of its base.

- state that the space displaee(! by a solid object is different fromlength or area, and requires a cubic unit of measure.

- use cubes (sugar, plastic, etc.) to measure volume.

VOCABULARY:

New terms: plane, solid, prism, cube, volume, displacement, space, face,edge

Review: rectangle, square, circle, parallelogram, parallel,dimension, triangle, polygon

CHALLENGE PROBLEM:

What is the difference between a map of the world and a globe?

DEVELOPMENT:

1. Use the challenge problem to discuss the difference between flator plane figures and solid figures. Have students recognize thatsolid figures are all around us. Although objects are solid figures,their faces maybe familiar plane figures. Have students list thenames of plan figures they have studied such as squares, rectangles,triangles, circles. Elicit that plane figures have only 2 dimensionsbut solid figures have 3 dimensions. Plane figures have sarfacc area

but solid figures occupy space.

2, Have students list examples from their experience of common solidobjects such as boxes, cans, balls, books, candles, buildings,dice, etc. One of the most common shapes is the rectangular solid.Which of the objects listed are in the shape of a rectangular solid(boxes, books, buildings)? Elicit the following description of arectangular solid: 6 rectangular surfaces called faces, each sideof a rectangle is an edge of the solid, the opposite faces areparallel and have the same size and shape. (See Figure 1 on next page.)

161

.:.

Figure 1

Have students make a F Jure of a rectangular solid and label theparts described.(Note: I shed lines indicate edges that cannot beseen.) Stress that th,, is a 2-dimensional representation of t,3-dimensional object.

3. A intangular solid is one type of prism. A prim is a solidfigure with faces that are polygons. One pair of these faces mustbe parall,-1 and nave the same size and shape. These are calledthe bases. The other faces must be parallelograms. (See Figure 2)

Figure 2

These are prisms: (1) (2)

These are not prisms: (5) (8) C......_.

(..--__=_.

Use Figure 2 to identify (.,), (2;, (3), and ( ,s pri3ms and(5), (6), (7), and (8) a' solids which are not 111:_sms. When thebases of a prism are rectang. 'le solid is called a rectangularsolid or rectangular prism (1). .nen the bases of a prism aretriangles the solid is called a triangular prism (2). Recall that

I .3

162 i C5

)

s

-70--

a triangular prism is used to break light into its componentcolors (the sprect-um). Elicit the reasons that (5), (6), (7),and (8) are not prisms. (For exa;,,ple in (' he bases are not thesame size and the other faces are not par elograms; in (8) thebases are not polygons.)

4. Recall that a plane figure has a surface area, but a solid figureoccupies space. How is the amount of space occupied by a solidfigure determ aed?

a. Review that to measure surface area, a unit which had a sur-face was chosen. This was called a square unit. Recall thereasons for this choice. To find the area was to determinethe number of square units in a surface.

b. Elicit that to measure space or volume, a unit must be chosenwhich has a volume. Whico of the objects shown in Figure 3could be used as q standard unit of volume? Why are the othersunsuitable?

(4)

Figure 3

(5)(6)

Name (5) as the best choice Elicit that it is a specialrectangular prism called a cube. Describe what is specialabout a cube. (Each face is a square and all edges are equal.

c. El it that to find the volume of an object means to determinethe number of cubes (cubic units cu in., cu ft, m3) contained

d.in the object.

Find the volume of the rectangular prism pictured in F'gure 4,or use a model of a rectangular prism and fill with un't cubes(if available).

Figure 4

5. Do applications.

163

iIPPLICATIONS:

A. Draw dashed lines in each of the sklids below to show the edges whichcannot be seen.

1. 2.

B. For each given prism name the shape of its base.

4.

-

/

164

1

,

)-- 1

C. Find the volume of each figure by counting the numbc of cubic units.

8.

3.

9.

6.

SUMMARY:

1. Hoy many faces does a rectangular solid have? How many edges?

2. A rectangular solid in which each face is a s;:.:are is called a

3. What does volume measure?

CUFF-HANGER:

What thethe volume of a rectangular prism whose dimensions are: length,41/2 in.,

width, 31/2 in.; height, 2 in.?

1 es

LESSON 2

AIM: To find volumes of rectangular solids


- Use the formula v = lwh to calcula,e the volume of a rectangular prism,where 1, w, and h are expressed as whole numbers, decimals and fractions.

- find a dimension of a rectangular prism if its volume and 2 dimensionsare given.

- appiy the formula v = lwh to the solution cf verbal problems.

'OCABULARY:

Review: length, width, height, rectangular prism, volume

CHALLENGE PROBLEM:

Find the volume of a rectaw-Ilar prism whose length equals 14 cm, widthequals 6 cm and height equi: s 8 cm.

DEVELOPMENT:

1. Discuss the challenge problem. Recall the procedure used to findthe volume of a rectangular prism (by counting the number of cubescontained in the prism).

3 cm 1

Z.Z. 2,74-T-Z777.727+' 777

,f j-,7" "="-7/74-/7/7i" , / /v

14 can

C:I1

8cm

a- Demonstrate that one method of filling the solid with cubesis to first fill the entire bottom with a single layer ofcm cubes. This bottom layer is called the base layer. HOY. canthe number of cubes in this base layer be found? Elicit thatthe prod..ct of the number of rows and the nurber of cm cubes ineach row gives the number of cubes in the base layer. Elicitthat the number of cubes in the base layer can be found byobtaining the product of the length a-d width of the base layer.Therefore, the number of cubes in the base layer equals 1 x w.

b. Pow many such layers will be contained in this recAngular

166 I CJ

prism? Elicit that the number of layers is equal to theheight of the rectangular prism. Therefore, the number oflayers = h.

c. Elicit that the total number of cm cubes contained in this rec-tangular solid can be found by multiplying the number of cubesin the base layer of the number of layers. Therefore the totalnumber of cubes in the rectangular prism(volume) is equal to itslength times its width times its height:

or V = lwh

2. Use the formula V = lwh to solve the challenge problem:

V= 14 x 6 x 8

V = 672

The volume is 672 cubic centimeters or 672 cm3

.

3. Similarly, the solution to the cliff-hanger problem can be found

by using: V = lwh

V= 41/2 x 31/2 x 2

V = 311/2

The volume is 311/2 cubic inches.

4. Do applications.

APPLICATIONS:

A. Complete the following table:

Volume Len th Width

1. 7 ft 9 ft

2. 8 in. 6 in.

3. 74 5 yd4

4. 6.1 m 2.3 m

S. 720 cm3 10 cm

6. 216 cu ft 6 ft

7. Syd Syd

Hei ht

8 ft

41/2 in.

61/2 yd

4.1 m

6 cm

8 ft

5 yd

B. Find the volume of a cube whose

1. edge is 5.2 cm.

2. edge is 11 in.

C. Find the volLme for each figure:

1.

1CIP-1

1- 0

3.

D. Solve each of the following problems:

1. A walk 40 ft long and 5 ft wide is to be paved. It is dug to adepth of 6 in. and filled with crushed rock. (Note: 6 in.= hft )

a. How many cu ft of crushed rock are nee,:ed?

b. Find the cost of the crushed rock at $2.35 per cu ft.

2. An aquarium is 44 in. long, 28 in. wide, and 15 in. high. Howmany gallons of water will the tank hold? (1 gallon = 231 cu in.)

3. The building code requires 200 cu ft of air per person in aclassroom. What is the maximum number of people who can occupy aclassroom 27 ft by 25 ft by 12 ft?

4. A liter is 1000 cm3 . Hoy many liters of water will fill a tank60 cm long, 30 cm wide anu 20 cm deep?

5. 41/2 cu ft of sand is poured into a sandbox and leveled to auniform depth. If the sandbox is 3 ft on each side, what is thedepth of the sand?

SUMMARY:

Review the formula for finding the volume of a rectangular prism.

CLIFF-HANGER:

Find the volume of the figure shown.

5 crn

166

10 cm

171

LESSON 3

AIM: To find the volumes of prisms and cylinders


- state tnat 1 x w is the area of the base of a rectangular prism,so that its volume is the product of the area of its base and itsheight.

- calculate the volume of any prism using the relationship that thevolume is equal to the area of the base times the height of the prism.

- distinguish between prisms and cylinders.

- find the volume of a (right-circular) cylinder.

VOCA:ULARY:

New terms: cylinder

Review: prism, volume

CHALLENGE PROBLEM:

Find the volume of a rectangular prism if its bas has air area of

84 cm2and its height is 8 cm.

DEVELOPMENT:

1. Discuss the challenge problem. Elicit that 84 cm2represents

the number of cubes in the base layer (84) (see development inLesson 2) and the height represents the number of layers (8).Therefore the volume of a rectangular prism can be found bymultiplying the number of cubes in each layer (area of baselayer) by the number of layers (height). Using B to represent

the area of the base (layer) write the formula foxy volume ofa rectangular prism: V = Bh (The volume is 672 cm..)

2. Elicit that the formula V = Bh is equivalent to the formulav = 1 wh whcn the base of a rectangular prism is a rectangle.The area of the base (B) is the length times the width (lw) .

Substituting B for lw, V = (lw)h becomes V = Bh.

(Note: In the formula V = Bh, h is the length of the edge between

the bases. The problems presented in this unit will deal onlywith right-circular cylinders and prisms, i.e., where the edgeis perpendicular to the bases.)

3. Recall the cliff-hanger problem. Name the figure (triangular prism).

Describe its base (a right triangle).

Figure on next page.

169

172

a. Since it has been established that the volume of a solid canbe found by multiplying the number of unit cubes in the baselayer (B) by the number of layers (h), the volume of thisprism may be found by V = Bh. In order to use this formula,it is necessary to calculate B (the area of the base).

b. Using the formula for the area of a riOt triangle, computethe area of this triangle to be:

A = Z x 8 x 5

A = 20 cm2

Therefore, the area of the base (B) = 20 cm2 and the volLmeof the triangular prism is:

V = 20 x 10

V = 200 cm3


S. Review the definition of prism as a solid whose bases are polygons.What is the name of a solid whose bases are circles? Name somecommon objects whose shape is a cyline..r. Have students suggest atechniqur. which could be used to find the volume of a cylinder(V = Bh). Use this formula to find the volume of a can 3 in.across and 6 in. high. Since the base of the cylinder is a circ12,its diameter is 3 in. The a-ea of a circle is found using A .

Therefore, since r = 1.5 in.

170

173

A = 3.14 x (1.5)2

A = 7.065 sq in.

The area of the base is 7.065 sq in.

The volume of the cylinder is found using V = Bh, therefore

V = 7.065 x 6

V = 42.39 cu in.

The volume of the can is 42.39 cu in.


APPLICATIONS:

A. Find the volume of each prism where B is the area of the base andh is the height.

11. B = 9 sq ft, h = 63 ft

2. B = 14 cm2

, h = 3.5 cm

3. B = 8.6 sq in.h = 1.4 in.1 1

i4. B = 3-7sq in. h = 2-T n.

B. Find the volume of each triangular prism.

2.

h 2_: 7'

---ri 7).13s ::---

4.

171i 74

C. Find the volume of each circular cylinder.

D. 1 Which container holds more, one that is 3 cm in diameter and 4 cmhigh or one that is 4 cm in diameter and 3 cm high? How manycubic centimers more can be held b; the container with the greatervolume?

2 The water tower on top of the Hillcrest Apartments is a cylinder14 feet across and 20 feet high. Find the volume of water inthe water tower when it is full. If there are 7.5 gallons ofwater per cubic foot, find the number of gallons in the tank.

3. A cylindrical eyedropper is .8 cm in diameter and 3.2 cm long. Howmany milliliters of liquid will the eyedropper hold (1 cm3 = 1 ml)?

4. A water tank truck is 10 feet in diameter and 42 feet long. It isbrought to a park to fill a wading pool 30 feet by 50 feet to a

172

15

depth of 2 feet. Will there be enough water? How much more water

is needed or how much extra water is left over?

SUMMARY:

Elicit that the formula V = Bh may be used to find the volume of a cube,a rectangular pri=, a triangular prism, and a cylinder.

CLIFF-HANGER:

If portions of french fries come in the two sizes indicated (see Figure 5),which do you think is the better buy? Explain your answer.

Figure 5

173

1;6

LESSON 4

AIM: To find the volumes of pyramids and cones


- distinguish between pyramids and prisms by recognizing that pyramidshare triangular sides and prisms have sides that are parallelograms.

- use V = 1 wh, or V = -.

17 the area of the base x height, to calculate the

volume of a pyramid.

- distinguish between cones (circular base) and pyramids (polygon base).

use V =1area of base x height to calculate the volume of a cone.

- identify a given geometric solid as a prism, a cylinder, a pyramid,or a cone.

VOCABULARY:

New terms: pyramid, cone

Review: prism, base, height, volume, cylinder, perpendicular

CHALLENGE PROBLO1:

If french fries come in the two serving sizes indicated (see Fig. 1),describe an experiment to show which is the better buy.

Figure 1

DEVELOPMENT:

1. Discuss the challenge problem. Elicit from students that theycould count or weigh the fries in each bag to show which is thebetter buy. This problem is equivalent to finding which containerhas the greater volume and comparing these volumes. Have studentsobserve that the heights and bases of both containers are the same;yet the volumes differ.

2. Elicit that the shape of the bag of french fries is a rectangularprism. Recall the characteristics of a rectangular prism. Halve

174

1.77

students identify the shape of the other container as a pyramid.Where have they seen this shape before? List the characteristicsof a pyramid: only one base, the base is a polygon, the facesare triangles. (Note: A pyramid with a square or rectangularbase is the most common, but pyramids may have any polygon as abase.)

3. The following experiment can be used to show the relationshipbetween the volume of a pyramid and that of a rectangular prismhaving the same base and height. (Models of these shapes arecommercially available for this purpose.) Fill the pyramid withsand or water and empty it into the pr'sm. Have students observethe height reached on the first filling. Repeat this procedureuntil the prism is full. This should take three fillings. Havestudents state the following in words: The volume of the rec-tangular prism is 3 times the volume of the pyramid or the volumeof a pyramid is one-third the volume of a rectangular prism ifboth have the same base area and height. Thus in the challengeproblem neither is the better buy since the unit price is the same.

4. Use the relationship just established to write the followingformula for the volume of a pyramid:

V =3 Bh

Students should recognize that B is the area of the base and h isthe length of the perpendicular drawn from the highest point tothe base.

5. Have students find the volume of the pyramid shown in Figure 2.

V = 1 BhFigure 2

The area of the base is 10 x 10 or 100 sc, in.; therefore usingthe formula:

v = T Bh

" 100 444

V= x x

1

V = 400 cu in.


78175

7 Have students imagine the shape of a pyramid if its base werecirc lar. Elicit that this is called a cone. List the charaz-teristics of a cone: only one base, the base is a circle. Whatare some objects which have the shape of a cone? What solid isclosely related to the cone? (Cylinder.) Have students hypothesizethe relatiolship between the volume of a cylinder and the vol-meof a cone if both have the same base radius (or diame ar) andheight. Perform an experiment similar to the one described inItem 3 to show that the volume of a cone is one -third the volumeof a cylinder if both have the same base radius and height, orV = 1 Bh.

3

8. Have students find the volume of the cone shown in Figure S.1

V =3

Bh Figure 3

The a_ ea of the base is A =11-r2 or

A = 3.14 x 10 x 10

A = 314 sq in,

1Therefore using the formula V = T Ph

9. Do Applications B pnd C.

APPIICATIONS:

A. Find the volume of each figure.

, 1 314 .1-24

v x T x1

V = 1256 in.

176

1 I 0

B. Find the volume of each figure. (See Figure 5)

Figure 5

(Js(,:.3 14 fO a.for a.

72)

Use 3.31 fo:

C. 1. A water tower is in the -hape of a cylinder surmounted by a cone(see Figure 6).

Figure 6

1^..1

;2!

a. Find the capacity of the water tower in cubic feet.

b. If there are 7.5 gallons of water in a cubic foot, find thenumber of gallons of water in the tank when the tank is 58%full.

2. 4, solid stone monument is in the shape of a rectangular prism_apped by a pyramid (see Fig. 7). If the stone weighs 173 poundsper cubic foot, find the weight of the monument.

Figure 7

9-

177 o

3

SUMMARY:

Have students identifyleach solid pictured and wr ' the formula for itsvolume (V = Bh or V = .- Bh).

1.

2.

CLIFF-HANGER:

,/----\....______,/

.........-----7----- /

-------

\----------

4.

5.

6.

A crystal ball is being shipped in a box. If the crystal ball is 10inches in diameter and just fits inside the box, how many cubic inchesof styrofoam are needed to fill the space left inside the box?

LESSON 5

AIM: To find volumes of spheres


- distinguish the sphere from other solid figures.

- use the formula V =4

'fl r3to find volumes of spheres.

VOCABULARY:

New terms: sphere, hemisphere

Review: volume, cylinder, cube

CHALLENGE PROBT.EM:

If a crystal ball 10 inches in diameter just fits inside a box, findthe volume of the box.

DEVELOPMENT:

1. Elicit from the class that the shape of the box must be a cubebecause of the properties of the ball. Since the ball is 10 incheswide, the length, width and height of the cube must be 10 inches.Therefore using V = Bh, obtain V = 10 x 10 x 10 or 1000 cu in. forthe volume of the cube.

2. Elicit that the geometric name for the ball is a sphere. Havestudents distinguish -Cie sphere from other solid figures (spherehas no plane surfaces). Name common objects shaped like a sphere.

3. Have students recognize that the volume of a sphere with a diameter = luis less than the volume of the cube whose length is 10 and can be fount by

the frllowing formula: 4 3V sfrr

4. Use the formula to find the volume of the crystal ball from th:schallenge problem.

4V = Tirr

3

V =4

(3.14) (53

)

V= 4x 3.14 x 5 x 5 x 5

V1570.00 15703 3

V = 523 cu in.

5, Solve the cliff-hanger problem. Since the volume of the cube is

.1000 cu in. and the volume of the ball is 523.1cu in., then the

1number of cubic inches left for the styrofo-m is 1000 - 5233 =

9

476= cu in.

6. Do applications.

179

APPLICATIONS:

A. Find the volume of each sphere described.

1. r = 7 in. (use Tr =

2. r = 1.3 m (use tr = 3.14)

3. d = 2 cm (use it = 3.14)

B. 1. A gas Ltorage tank is in the shape of a sphere. The tank is 30 ftin diameter. Now many cubic feet of gas can be stored in this tank?

2. A laser light show is presented in a hemisphere (hemisphere is 1/2of a sphere) 42 ft in diameter. If 100 cl ft of space is requiredfor each person occupying the room, what the maximum occupancyfor tlis room?

3. Tennis balls are vacuum packed in cylindrical cans, 3 balls to thecan. If each tennis ball is approximately 2 inches in diameter,how much air must be removed from the can before it is sealed atthe factory?

SUMMARY:

1. Name each of the geometric figures represented by the followingobjects.

2. State the formula .sed to find the volume of each object.

a. a base all

b. a teepee

c. a. crate of apples

d. a water pipe

e. the tol of the Washington Monument

180

Teacher's notes

181

It is the policy of the New York City Board of Education not liscrimmate on the has of race, creed, national origin, age,handicapping condition, or sex in its educational programs, ac-tivives, and employment policies, as required by law In-quiries regarding compliance with appropriate laws may be directed to Mercedes NPsfield, Director, Office of Equal Op-portunky, 110 Livingston Street, Brooklyn, New Yr l< 11201; or to Charles Teiada, Director, Office of 'Ay!! Rights, Depart-ment of Education, 26 Federal Plaza, Room 33.1s0, New York, N1 10278.

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